The Flat Earth Society
Flat Earth Discussion Boards => Flat Earth Investigations => Topic started by: garygreen on October 22, 2018, 04:07:35 AM
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You think a spinning earth would add additional pull straight down towards its surface? ???
Make a thread in the discussion forums about your idea please. I am curious to know what you think the situation is.
this situation already appears in many physics textbooks. here is an example:
(https://i.imgur.com/POP5CAZ.png)
equation 11.6 has two terms: one for the "straight down" acceleration, and one for the "sideways" acceleration. at 45 deg north, the magnitude of the "straight down" acceleration is 9.8 m/s2. the magnitude of the "sideways" acceleration is 0.0169 m/s2. the sideways acceleration is tiny compared to the downward acceleration.
by F=ma, an 80kg human would experience a "sideways" force of ~1.3 N. that's only ~2x the force required to press a key on a keyboard. but it's not localized like when you press a key with your finger; it's distributed throughout the whole volume of your body.
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You seem to be quote-mining in desperation.
If a static planet were put into rotation, why would it cause additional pull directly downwards towards its surface? Please explain for us in your own words.
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The equation F=ma doesn't universally use kilograms.
The value you got of 0.0212 m/s^2 is 21.2 mm / s^2, which is close to the value that the person in the video got.
That is equivalent to being pulled horizontally with weight of 30 quarters on a 175lb person.
This means that, if we were facing in one direction and being pulled to our left by 30 quarters, that we could turn around by 180 degrees and be pulled to our right by 30 quarters. Please tell us, in a thread in the appropriate forum, why we shouldn't be able to feel this or detect it with experiment.
When you lay it out like that it sure sounds hard to believe doesn't it? But how much is it really on one part of you? Our 175lb person is about 2.5 cubic feet of volume, or 4320 cubic inches. So every cubic inch is being pulled differently by the weight of about 0.08 grams. So the palm of your hand will in theory feel a shift equivalent to less than a tenth the weight of a paperclip (which is approx 1 gram). Why do you expect to be able to feel this again? If someone places a paperclip on your shoulder, do you feel the extra weight? I suspect you might feel how it brushes the hair along your arm, but expect a fair bit of doubt if you claim you can tell when someone places a paperclip on your shoulder from the added weight alone, much less something 1/10 of its weight.
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The equation F=ma doesn't universally use kilograms.
The value you got of 0.0212 m/s^2 is 21.2 mm / s^2, which is close to the value that the person in the video got.
That is equivalent to being pulled horizontally with weight of 30 quarters on a 175lb person.
This means that, if we were facing in one direction and being pulled to our left by 30 quarters, that we could turn around by 180 degrees and be pulled to our right by 30 quarters. Please tell us, in a thread in the appropriate forum, why we shouldn't be able to feel this or detect it with experiment.
Ok. The SI unit of mass IS the kilogram, that is just a definition. I guess you can get the same answer in different units if you convert it all but it’s a bit of a clumsy way of working.
Anyway, that rate of acceleration is equivalent to being in a car which is going from 0-60mph in about 21 minutes. I personally doubt you’d be able to feel that, but I cannot prove that.
But, and this is the thing you keep ignoring, MY MATHS IS WRONG. It’s wrong for the same reason the maths in the video is wrong. I have assumed that you are going at 460m/s in a STRAIGHT LINE in one direction and, over 12 hours you change velocity so that you are now going at 460m/s in the opposite direction. That is the logic of the video. But that is not what the heliocentric model claims. It claims we are going in a circle at a constant angular velocity. Because you are moving in a circle, not a straight line, you need to use different maths. I gave you a link in the other thread to the correct maths. That rotation DOES exert a force and you DO weigh a different amount at the equator but:
1) You don’t weigh different amounts during the course of the day, the angular velocity due to the rotation is constant thus the force on the body is.
2) The difference in your weight between the equator and the poles is 0.3%. The maths is in the other thread. If an 180lb person got on a plane at the pole and got off at the equator then their weight would have changed by little over half a pound. If you put on a pound or lose a pound are you seriously claiming you feel noticeably different?
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The equation F=ma doesn't universally use kilograms.
The value you got of 0.0212 m/s^2 is 21.2 mm / s^2, which is close to the value that the person in the video got.
That is equivalent to being pulled horizontally with weight of 30 quarters on a 175lb person.
This means that, if we were facing in one direction and being pulled to our left by 30 quarters, that we could turn around by 180 degrees and be pulled to our right by 30 quarters. Please tell us, in a thread in the appropriate forum, why we shouldn't be able to feel this or detect it with experiment.
When you lay it out like that it sure sounds hard to believe doesn't it? But how much is it really on one part of you? Our 175lb person is about 2.5 cubic feet of volume, or 4320 cubic inches. So every cubic inch is being pulled differently by the weight of about 0.08 grams. So the palm of your hand will in theory feel a shift equivalent to less than a tenth the weight of a paperclip (which is approx 1 gram). Why do you expect to be able to feel this again? If someone places a paperclip on your shoulder, do you feel the extra weight? I suspect you might feel how it brushes the hair along your arm, but expect a fair bit of doubt if you claim you can tell when someone places a paperclip on your shoulder from the added weight alone, much less something 1/10 of its weight.
One dollar bill weighs one gram. Cut out a tenth of that, fold it, and hold it in you hand? Can you feel the weight?
Now, every square inch of your body should also feel that weight, pulling you horizontally. When you turn one way you should feel it pulling in one direction, and then when you turn around you should feel it pulling in the other direction. We should instantaneously feel this deflection. Yet we do not feel this at all.
But, and this is the thing you keep ignoring, MY MATHS IS WRONG. It’s wrong for the same reason the maths in the video is wrong. I have assumed that you are going at 460m/s in a STRAIGHT LINE in one direction and, over 12 hours you change velocity so that you are now going at 460m/s in the opposite direction. That is the logic of the video. But that is not what the heliocentric model claims. It claims we are going in a circle at a constant angular velocity. Because you are moving in a circle, not a straight line, you need to use different maths. I gave you a link in the other thread to the correct maths. That rotation DOES exert a force and you DO weigh a different amount at the equator but:
1) You don’t weigh different amounts during the course of the day, the angular velocity due to the rotation is constant thus the force on the body is.
2) The difference in your weight between the equator and the poles is 0.3%. The maths is in the other thread. If an 180lb person got on a plane at the pole and got off at the equator then their weight would have changed by little over half a pound. If you put on a pound or lose a pound are you seriously claiming you feel noticeably different?
If that pound is trying to fall horizontally, while the rest of you fell vertically, why wouldn't you feel noticeably different? That is an imbalance, and should be detectable.
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The equation F=ma doesn't universally use kilograms.
The value you got of 0.0212 m/s^2 is 21.2 mm / s^2, which is close to the value that the person in the video got.
That is equivalent to being pulled horizontally with weight of 30 quarters on a 175lb person.
This means that, if we were facing in one direction and being pulled to our left by 30 quarters, that we could turn around by 180 degrees and be pulled to our right by 30 quarters. Please tell us, in a thread in the appropriate forum, why we shouldn't be able to feel this or detect it with experiment.
When you lay it out like that it sure sounds hard to believe doesn't it? But how much is it really on one part of you? Our 175lb person is about 2.5 cubic feet of volume, or 4320 cubic inches. So every cubic inch is being pulled differently by the weight of about 0.08 grams. So the palm of your hand will in theory feel a shift equivalent to less than a tenth the weight of a paperclip (which is approx 1 gram). Why do you expect to be able to feel this again? If someone places a paperclip on your shoulder, do you feel the extra weight? I suspect you might feel how it brushes the hair along your arm, but expect a fair bit of doubt if you claim you can tell when someone places a paperclip on your shoulder from the added weight alone, much less something 1/10 of its weight.
One dollar bill weighs one gram. Cut out a tenth of that, fold it, and hold it in you hand? Can you feel the weight?
Now, every square inch of your body should also feel that weight, pulling you horizontally. When you turn one way you should feel it pulling in one direction, and then when you turn around you should feel it pulling in the other direction. Yet we do not feel this distributed weight at all.
I definitely cannot feel the 'weight' of a dollar bill, much less a fraction of it. Can I feel on in my hand? Yup. My sense of touch is working there. But if I put a glove on and add a dollar bill to it? I have no idea that bill is there. I would also personally call BS if you were to claim that as well based on my own experiences. Go try it yourself though. Put on a glove. Lay a dollar bill on your hand. Or better yet, hold out your gloved hand and have someone else place it on there. See if you can guess better than 50% of the time whether they placed it on there or not. I'd wager money you can't approach that sort of accuracy unless they accidentally brush your hand or something.
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I often don't feel the weight of my clothes.
I am trying REALLY hard to feel the weight of my vest... I can't.
If I spin a lot, the vest's loose bits (yes i'm not that fat) seem to pull away from me. I think that may not be the effect you're hoping for though.
How would I test this tiny shift in weight without moving?... I actually can't think of a way.
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I definitely cannot feel the 'weight' of a dollar bill, much less a fraction of it. Can I feel on in my hand? Yup. My sense of touch is working there. But if I put a glove on and add a dollar bill to it? I have no idea that bill is there. I would also personally call BS if you were to claim that as well based on my own experiences. Go try it yourself though. Put on a glove. Lay a dollar bill on your hand. Or better yet, hold out your gloved hand and have someone else place it on there. See if you can guess better than 50% of the time whether they placed it on there or not. I'd wager money you can't approach that sort of accuracy unless they accidentally brush your hand or something.
It's not just one dollar. It's more like the weight of 174 dollars, distributed all over your body. When you turn one direction you should feel it pulling in one direction, and when you turn 180 degrees around, you feel it pulling in the other direction. We should instantly be able to turn this on and off, and reverse the effect, by changing directions. Yet this is not felt at all.
The argument of "well if you are wearing clothes..." is irrelevant since it affects all parts of your body and your clothes are resting on you being pulled downwards while your body is being pulled horizontally.
174 dollars, or 174.147246 grams, is a significant amount of weight to shift around and should absolutely be noticeable.
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It's not just one dollar. It's more like the weight of 174 dollars, distributed all over your body.
That's 6 ounces "distributed all over your body". I fail to see how I would notice it. If you smeared six ounces of butter all over yourself would you feel any weight distribution change? I think not.
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It's not just one dollar. It's more like the weight of 174 dollars, distributed all over your body.
That's 6 ounces "distributed all over your body". I fail to see how I would notice it. If you smeared six ounces of butter all over yourself would you feel any weight distribution change? I think not.
This is horizontal weight, which imbalances you, and can be reversed by changing directions, not vertical weight.
If you had a button which reversed its weight on your body from 6 ounces to -6 ounces whenever you pressed it, would you not feel it?
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If you had a button which reversed its weight on your body from 6 ounces to -6 ounces whenever you pressed it, would you not feel it?
I think I would, as long as it happened pretty suddenly.
Otherwise, the vertical weight of the button would be more noticeable.
I am assuming this is a big button? I have not been able to follow the maths so far and examples have varied from a paper clip to paper money.
A better example might be a fly on a string, it goes this way and that without having to account for it’s verticality. The dickens.
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I definitely cannot feel the 'weight' of a dollar bill, much less a fraction of it. Can I feel on in my hand? Yup. My sense of touch is working there. But if I put a glove on and add a dollar bill to it? I have no idea that bill is there. I would also personally call BS if you were to claim that as well based on my own experiences. Go try it yourself though. Put on a glove. Lay a dollar bill on your hand. Or better yet, hold out your gloved hand and have someone else place it on there. See if you can guess better than 50% of the time whether they placed it on there or not. I'd wager money you can't approach that sort of accuracy unless they accidentally brush your hand or something.
It's not just one dollar. It's more like the weight of 174 dollars, distributed all over your body. When you turn one direction you should feel it pulling in one direction, and when you turn 180 degrees around, you feel it pulling in the other direction. We should instantly be able to turn this on and off, and reverse the effect, by changing directions. Yet this is not felt at all.
The argument of "well if you are wearing clothes..." is irrelevant since it affects all parts of your body and your clothes are resting on you being pulled downwards while your body is being pulled horizontally.
174 dollars, or 174.147246 grams, is a significant amount of weight to shift around and should absolutely be noticeable.
Can you prove it? The math indicates laying a dollar on your hand would be roughly equivalent if you can feel your hand change in weight. Might need to use two dollars. Your entire argument right now "Well you totally should!" and we can go about this in circles forever. Test it. Give us an experiment you feel better reflects things than the dollar experiment I outlined above. But remember, anything that you do has to adhere to roughly 1 gram per cubic inch. You can't just put on a backpack and throw 174 grams of something in there and say you can feel it. You've just focused those 174 grams into the area of the straps on your back. Hence my original suggestion of a dollar bill (or two) onto your gloved hand. This controls about as many points as I can think of offhand. It ensures you aren't just using your sense of touch to know it's there by blocking it off with a glove. It localizes the effect to an area about equivalent to our 1gr/inch^3 above. The blindfold should work to reduce or remove the 'ghost' feeling if you can see them being placed on your hand. Lastly having someone else place them on your hand at a random time, helps to reduce the same 'ghost' feeling that you could get by placing them on your hand on your own. If you can successfully 'know' that they've placed it on your hand a percentage of the time greater than what just guessing would get you (a standard deviation or so above 50%) you will have provided evidence to substantiate your claim that you should be able to feel this change in force/weight. If you can't, it's evidence your claim that we should be able to feel it is false. Science!
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You seem to be quote-mining in desperation.
i don't think you know what quote-mining is. this is just a standard treatment of centrifugal force from a popular graduate textbook on mechanics. what specifically do you take issue with?
If a static planet were put into rotation, why would it cause additional pull directly downwards towards its surface? Please explain for us in your own words.
it wouldn't. the centrifugal force acts radially outward from the axis of rotation. this means that someone standing at 45 deg north experiences an additional "sideways" force (the quarters analogy you keep making). the equation (11.6) from my textbook tells you how to calculate that force. the r-hat term is the downward force, and the theta-hat term is the "sideways" force.
the magnitude of the "sideways" acceleration is 0.0169 m/s2. why is it not 0.0212 m/s2? because not all of the centrifugal force is in the "sideways" direction. some of it is in the "upward" direction.
(https://i.imgur.com/i681a4R.png)(https://i.imgur.com/mLmWLmC.png)(https://i.imgur.com/Kw0An1V.png)
again, this tiny acceleration is distributed evenly throughout your entire volume.
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I definitely cannot feel the 'weight' of a dollar bill, much less a fraction of it. Can I feel on in my hand? Yup. My sense of touch is working there. But if I put a glove on and add a dollar bill to it? I have no idea that bill is there. I would also personally call BS if you were to claim that as well based on my own experiences. Go try it yourself though. Put on a glove. Lay a dollar bill on your hand. Or better yet, hold out your gloved hand and have someone else place it on there. See if you can guess better than 50% of the time whether they placed it on there or not. I'd wager money you can't approach that sort of accuracy unless they accidentally brush your hand or something.
It's not just one dollar. It's more like the weight of 174 dollars, distributed all over your body. When you turn one direction you should feel it pulling in one direction, and when you turn 180 degrees around, you feel it pulling in the other direction. We should instantly be able to turn this on and off, and reverse the effect, by changing directions. Yet this is not felt at all.
The argument of "well if you are wearing clothes..." is irrelevant since it affects all parts of your body and your clothes are resting on you being pulled downwards while your body is being pulled horizontally.
174 dollars, or 174.147246 grams, is a significant amount of weight to shift around and should absolutely be noticeable.
Can you prove it?
https://www.quora.com/What-is-the-minimum-amount-of-acceleration-that-a-human-can-detect
What is the minimum amount of acceleration that a human can detect?
A2A: There are many research papers written on this. If you want greater depth of details, I would suggest doing a search on "treshold of perception for vibration". As it happens, I did my thesis on a topic along these lines...
First, there isn't one set number. Second, much of the focus on this is on vibrational accelerations--something you are sensing "moving". So this is not necessarily applicable, for instance, to a gentle acceleration of an elevator. For something like an elevator, I would speculate somewhere around 0.1 - 0.5 m/sec^2 would be the threshold, i.e. 0.01 - 0.05g. Often it is the "jerk" that we notice in a situation like this (rather than the constant rate of acceleration itself).
I would say there are 3-4 major variables in-play:
1) Frequency of the vibration (assuming a sinusoidal pulsation)
2) Body part sensing the vibration (feet/hands/seat/whole body)
3) Axis of vibration (fore-aft/lateral/up-down).
4) Length of the vibration
Generally these studies focus on a range of ~2Hz - 200 Hz. Within this range, the ballpark number for sensitivity threshold is ~0.01 - 0.10 m/s^2 (so roughly 0.001 - 0.010 g).
Taking his advice on searching for "threshold of perception for vibration" I got the following:
https://books.google.com/books?id=-_JV63L6RMcC&lpg=PA262&ots=vNfENVZrtY&dq=acceleration%20threshold%20sensation%20human%20%22m%20%2F%20s%20%5E%202%22&pg=PA262#v=onepage&q&f=false
(https://i.imgur.com/bO64a5h.png)
0.01 m^2 = 10 mm^2
0.02 m^2 = 20 mm^2
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If a static planet were put into rotation, why would it cause additional pull directly downwards towards its surface? Please explain for us in your own words.
it wouldn't.
You started this thread by trying to justify your assertion that a rotating earth would cause downwards acceleration. Now you say that it would not cause downwards acceleration?
There is nothing wrong with just admitting that you were wrong.
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Oh, awesome. So the Quora guy agrees completely as he gives 0.1 m/s^2 at the low end, and we're talking about a speed around 20% of that. The second source you link is very clear that it's talking about *vibrating* acceleration, which is not what's being discussed. Even if it were, it suggests our 0.02 m/s^2 is on the very lowest edge of human perception. Both of your sources appear to be in agreement that we should not be able to feel the acceleration due to the spin of the Earth. Even 'best case' suggests it would be likely difficult to do so, assuming a set of circumstances that aren't reality. Thank you for presenting a rather strong case against your own premise.
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Snipped for post brevity
Oh, awesome. So the Quora guy agrees completely as he gives 0.1 m/s^2 at the low end, and we're talking about a speed around 20% of that. The second source you link is very clear that it's talking about *vibrating* acceleration, which is not what's being discussed. Even if it were, it suggests our 0.02 m/s^2 is on the very lowest edge of human perception. Both of your sources appear to be in agreement that we should not be able to feel the acceleration due to the spin of the Earth. Even 'best case' suggests it would be likely difficult to do so, assuming a set of circumstances that aren't reality. Thank you for presenting a rather strong case against your own premise.
Ranges between 0.01 m/s^2 and 0.0221 m/s^2 are in the "probable perception" to "clear perception" range. Yet no one has perceived the rotation of the earth. Why?
The author of the quora article who wrote a thesis on the subject told us to look up vibrational perception. It makes sense. We are detecting different accelerations depending on which direction we face, and can go from minus to plus depending on one's position. Yet again, however, no one has felt this acceleration.
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Snipped for post brevity
Oh, awesome. So the Quora guy agrees completely as he gives 0.1 m/s^2 at the low end, and we're talking about a speed around 20% of that. The second source you link is very clear that it's talking about *vibrating* acceleration, which is not what's being discussed. Even if it were, it suggests our 0.02 m/s^2 is on the very lowest edge of human perception. Both of your sources appear to be in agreement that we should not be able to feel the acceleration due to the spin of the Earth. Even 'best case' suggests it would be likely difficult to do so, assuming a set of circumstances that aren't reality. Thank you for presenting a rather strong case against your own premise.
Ranges between 0.01 m/s^2 and 0.0221 m/s^2 are in the "probable perception" to "clear perception" range. Yet no one has perceived the rotation of the earth. Why?
The author of the quora article who wrote a thesis on the subject told us to look up vibrational perception. It makes sense. We are detecting different accelerations depending on which direction we face, and can go from minus to plus depending on one's position. Yet again, however, no one has felt this acceleration.
Those ranges are for something vibrating at speeds of 2-20Hz, and I would wager the low vibration end is towards the higher Hz end. He specifically calls out an elevator too, noting the lowest was around 0.1 m/s^2. The elevator is much closer to normal conditions on Earth.
But sure, lets go with just spinning, and we'll go with the best case scenario, looking for 0.0221 m/s^2 is detectable. But wait, we *also* have to account for the 'noise' we'll be making as we spin. That will be far more noticeable most likely. So if we go with our 80kg guy from before. The average length from hip to hip is about 3.6m. We'll cut down to 1.2m radius and 40kg. I think that should give us a rough sort of midpoint, but I'll run a few others too for fun. That gives us a speed of spin of roughly 15 m/s^2 for our 40k, generating centripetal force equal to 7500 Newtons. That's about 764 kg m/s^2 of force. Not sure anyone is gonna notice an extra 0.0221. Well let's see what we can do to get this a bit more favorable, shall we?
Hrmm, the lowest I'm getting, using the shortest person to have lived, gives me about 23.2 kg m/s^2. For whom I was being somewhat generous as well imo. Radius of 0.30m (about his 'center' using a similar method as above), gives us a velocity of 3.7 m/s^2, and a weight of 5kg. I'm not seeing what part of you is supposed to have felt this acceleration Tom, especially not feeling it over simply the force generated by spinning yourself around. Feel free to check my math if you like, I encourage it in fact. But your hypothesis that we should be able to feel it by spinning is looking dead in the water again.
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You started this thread by trying to justify your assertion that a rotating earth would cause downwards acceleration.
at no point has that ever been my argument. my argument has always been the exact opposite of that, hence the "you weigh less in ecuador" argument. the centrifugal force always acts radially away from the axis of rotation.
i'm not sure what your beef here is. the op addresses exactly the thing you are talking about with the quarters. you can calculate the "horizontal pulling" you describe using the expression shown. the magnitude of the "horizontal pulling" is 0.0169 m/s2.
if you want to talk about how noticeable that centrifugal force is, we can compare it to a merry-go-round. unlike vibration, that's an apples-to-apples comparison. recall that by F=ma, an 80kg human would experience a "horizontal pulling" force of ~1.3 N.
centrifugal force = object mass * radius of rotation * angular velocity 2. let's assume the same 80 kg person is riding on a merry-go-round with a 3 meter radius (typical size according to the internet). for that mass and radius, the angular velocity required to produce a centrifugal force of 1.3 N is 0.074 radians per second. that's slower than the second hand on a clock. the period of such a rotation is 85 seconds.
punchline: the "horizontal pulling" you experience at 45 deg north is equivalent to the horizontal pulling you would experience standing on a merry-go-round that takes 85 seconds to complete a single rotation.
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If you want to see all the effects of gravity just go to the 'International Gravity Formula'. Everything is outlined there. The effects of centrifugal force should be 0.03382 * Cos^2(Lat). This is just the difference between what you would experience if the earth were NOT rotating and if the earth was, at the normal 1 rotation per 24 hours. Of course the centrifugal force would just be a correction to the normal force of gravity because it's a force vector in exactly the opposite direction. The heliocentric force due to acceleration is just the combination of the earths rotation on it's own axis and the rotation around the sun. Unfortunately for the FET folks there are increasingly inexpensive MEMS devices out there that can actually measure accelerations on the surface of the earth and they come out with the expected values. These devices are similar to those found inside your cell phones only better and more sensitive. Right now the data from these devices can measure the accelerations due to the heliocentric speed changes but it takes a while. The acceleration data is buried in a lot of noise and you have to take measurements over a period of time, then filter the data points to actually extract the readings you want. This would mean you could easily take a device like this and take measurements at different latitudes. If you got anywhere near the expected values than it would prove that the earth was both a sphere and rotating. Of course there already is a network of instruments around the world that have already done this and they publish their results. The force of gravity changes constantly due to the fact that the earth is not static. Gravity is simply one center of mass attracting another. If the location of that center of mass changes then the force would also change. The network constantly measures the acceleration forces of gravity in many different locations around the world. I didn't expect to see this mountain of information coming my way, but I took out and dusted off my own dynamics book that I used in college years ago. All I have to do now is get myself back up to speed on the subject matter.
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Snipped for post brevity
Oh, awesome. So the Quora guy agrees completely as he gives 0.1 m/s^2 at the low end, and we're talking about a speed around 20% of that. The second source you link is very clear that it's talking about *vibrating* acceleration, which is not what's being discussed. Even if it were, it suggests our 0.02 m/s^2 is on the very lowest edge of human perception. Both of your sources appear to be in agreement that we should not be able to feel the acceleration due to the spin of the Earth. Even 'best case' suggests it would be likely difficult to do so, assuming a set of circumstances that aren't reality. Thank you for presenting a rather strong case against your own premise.
Ranges between 0.01 m/s^2 and 0.0221 m/s^2 are in the "probable perception" to "clear perception" range. Yet no one has perceived the rotation of the earth. Why?
The author of the quora article who wrote a thesis on the subject told us to look up vibrational perception. It makes sense. We are detecting different accelerations depending on which direction we face, and can go from minus to plus depending on one's position. Yet again, however, no one has felt this acceleration.
Those ranges are for something vibrating at speeds of 2-20Hz, and I would wager the low vibration end is towards the higher Hz end. He specifically calls out an elevator too, noting the lowest was around 0.1 m/s^2. The elevator is much closer to normal conditions on Earth.
But sure, lets go with just spinning, and we'll go with the best case scenario, looking for 0.0221 m/s^2 is detectable. But wait, we *also* have to account for the 'noise' we'll be making as we spin. That will be far more noticeable most likely. So if we go with our 80kg guy from before. The average length from hip to hip is about 3.6m. We'll cut down to 1.2m radius and 40kg. I think that should give us a rough sort of midpoint, but I'll run a few others too for fun. That gives us a speed of spin of roughly 15 m/s^2 for our 40k, generating centripetal force equal to 7500 Newtons. That's about 764 kg m/s^2 of force. Not sure anyone is gonna notice an extra 0.0221. Well let's see what we can do to get this a bit more favorable, shall we?
Hrmm, the lowest I'm getting, using the shortest person to have lived, gives me about 23.2 kg m/s^2. For whom I was being somewhat generous as well imo. Radius of 0.30m (about his 'center' using a similar method as above), gives us a velocity of 3.7 m/s^2, and a weight of 5kg. I'm not seeing what part of you is supposed to have felt this acceleration Tom, especially not feeling it over simply the force generated by spinning yourself around. Feel free to check my math if you like, I encourage it in fact. But your hypothesis that we should be able to feel it by spinning is looking dead in the water again.
People don't just rotate on their own power. People spin in cars, aircraft, in freefall. No one has detected this vibration as they are turned in orientation. If this were a thing, we would know about it.
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Snipped for post brevity
Oh, awesome. So the Quora guy agrees completely as he gives 0.1 m/s^2 at the low end, and we're talking about a speed around 20% of that. The second source you link is very clear that it's talking about *vibrating* acceleration, which is not what's being discussed. Even if it were, it suggests our 0.02 m/s^2 is on the very lowest edge of human perception. Both of your sources appear to be in agreement that we should not be able to feel the acceleration due to the spin of the Earth. Even 'best case' suggests it would be likely difficult to do so, assuming a set of circumstances that aren't reality. Thank you for presenting a rather strong case against your own premise.
Ranges between 0.01 m/s^2 and 0.0221 m/s^2 are in the "probable perception" to "clear perception" range. Yet no one has perceived the rotation of the earth. Why?
The author of the quora article who wrote a thesis on the subject told us to look up vibrational perception. It makes sense. We are detecting different accelerations depending on which direction we face, and can go from minus to plus depending on one's position. Yet again, however, no one has felt this acceleration.
Those ranges are for something vibrating at speeds of 2-20Hz, and I would wager the low vibration end is towards the higher Hz end. He specifically calls out an elevator too, noting the lowest was around 0.1 m/s^2. The elevator is much closer to normal conditions on Earth.
But sure, lets go with just spinning, and we'll go with the best case scenario, looking for 0.0221 m/s^2 is detectable. But wait, we *also* have to account for the 'noise' we'll be making as we spin. That will be far more noticeable most likely. So if we go with our 80kg guy from before. The average length from hip to hip is about 3.6m. We'll cut down to 1.2m radius and 40kg. I think that should give us a rough sort of midpoint, but I'll run a few others too for fun. That gives us a speed of spin of roughly 15 m/s^2 for our 40k, generating centripetal force equal to 7500 Newtons. That's about 764 kg m/s^2 of force. Not sure anyone is gonna notice an extra 0.0221. Well let's see what we can do to get this a bit more favorable, shall we?
Hrmm, the lowest I'm getting, using the shortest person to have lived, gives me about 23.2 kg m/s^2. For whom I was being somewhat generous as well imo. Radius of 0.30m (about his 'center' using a similar method as above), gives us a velocity of 3.7 m/s^2, and a weight of 5kg. I'm not seeing what part of you is supposed to have felt this acceleration Tom, especially not feeling it over simply the force generated by spinning yourself around. Feel free to check my math if you like, I encourage it in fact. But your hypothesis that we should be able to feel it by spinning is looking dead in the water again.
People don't just rotate on their own power. People spin in cars, aircraft, in freefall. No one has detected this vibration as they are turned in orientation. If this were a thing, we would know about it.
So you have no actual refutation, simply repeating "We would know" over and over. I just laid out the math for why we *wouldn't* notice this 'vibration' due to the Earth's rotation. It's incredibly tiny compared to the forces simply spinning would have upon your body, regardless of just *how* you're spinning. If I spin you at 20Hz you'll experience even more force from that motion. You have yet to provide a shred of credible evidence that we should be able to feel the 0.01-0.0221 m/s^2 of additional acceleration from the spinning of the Earth, compared to the accelerations we would be feeling from other sources in our attempt to sense this one. 'We would know' is neither empirical, nor zetetic. Prove it. All you've provided so far shows we just might be able to sense it under the right circumstances. But every circumstance you've suggested to attempt to sense it results in far greater noise than a human can possibly ignore. So far all I've gotten is even more that there's no reason to think we could detect the spinning/rotation of the Earth on our own.
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But, and this is the thing you keep ignoring, MY MATHS IS WRONG. It’s wrong for the same reason the maths in the video is wrong. I have assumed that you are going at 460m/s in a STRAIGHT LINE in one direction and, over 12 hours you change velocity so that you are now going at 460m/s in the opposite direction. That is the logic of the video.
That's the error in the video, it only shows acceleration in "x-direction". What about the "y-direction"? On the 6am and 6pm positions, you would have to place the same arrows as on 12am and 12pm, but in perpendicular direction. This guy, who produced the video, suppressed this acceleration. If you now combine these vectors - decreasing with a cos-function, increasing with a sin-function - the resulting vector would be radial from (or to) earth center.
You could have it easier, if you consider the setup as this what it is: (sorry I don't know the exact term in English) a steady circular motion: The only force or acceleration produced by this motion is centripetal/centrifugal force:
a) radial to the center
b) constant
So at the equator the force is anti-parallel to gravity. The only thing you could measure - not notice - is that things at the equator are a bit less heavy, and this doesn't change in the course of a day (constant!). And most notable there's no additional acceleration in the direction along the curved path.
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So you have no actual refutation, simply repeating "We would know" over and over. I just laid out the math for why we *wouldn't* notice this 'vibration' due to the Earth's rotation. It's incredibly tiny compared to the forces simply spinning would have upon your body, regardless of just *how* you're spinning. If I spin you at 20Hz you'll experience even more force from that motion. You have yet to provide a shred of credible evidence that we should be able to feel the 0.01-0.0221 m/s^2 of additional acceleration from the spinning of the Earth, compared to the accelerations we would be feeling from other sources in our attempt to sense this one. 'We would know' is neither empirical, nor zetetic. Prove it. All you've provided so far shows we just might be able to sense it under the right circumstances. But every circumstance you've suggested to attempt to sense it results in far greater noise than a human can possibly ignore. So far all I've gotten is even more that there's no reason to think we could detect the spinning/rotation of the Earth on our own.
There was already plenty of noise in the experiments in which it was determined that people can feel 0.01+ m/s^2. Bodily movements, attention, etc. Yet 0.01+ m/s^2 was still set as the threshold for human perception.
You are the one who needs to prove that there is some mysterious noise that prevents one from detecting such acceleration. You are claiming that if we are turning in one direction, that there is too much noise to detect vibration. What noise? Are you claiming that the human body's ability to detect vibration from any and all directions turns off if one is turning? Are you claiming that when we turn, that the human body exactly cancels out any and all vibrations from all directions?
My part is already done. You are the one tying to bring in mysterious variables here, in increasingly desperate tactics to claim that it is all too noisy to detect. You are the one who has no actual refutation, instead resting your argument on mysterious noise. Substantiate your mysterious variables.
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So you have no actual refutation, simply repeating "We would know" over and over. I just laid out the math for why we *wouldn't* notice this 'vibration' due to the Earth's rotation. It's incredibly tiny compared to the forces simply spinning would have upon your body, regardless of just *how* you're spinning. If I spin you at 20Hz you'll experience even more force from that motion. You have yet to provide a shred of credible evidence that we should be able to feel the 0.01-0.0221 m/s^2 of additional acceleration from the spinning of the Earth, compared to the accelerations we would be feeling from other sources in our attempt to sense this one. 'We would know' is neither empirical, nor zetetic. Prove it. All you've provided so far shows we just might be able to sense it under the right circumstances. But every circumstance you've suggested to attempt to sense it results in far greater noise than a human can possibly ignore. So far all I've gotten is even more that there's no reason to think we could detect the spinning/rotation of the Earth on our own.
There was already plenty of noise in the experiments in which it was determined that people can feel 0.01+ m/s^2. Bodily movements, attention, etc. Yet 0.01+ m/s^2 was still set as the threshold for human perception.
You are the one who needs to prove that there is some mysterious noise that prevents one from detecting such acceleration. You are claiming that if we are turning in one direction, that there is too much noise to detect vibration. What noise? Are you claiming that the human body's ability to detect vibration from any and all directions turns off if one is turning? Are you claiming that when we turn, that the human body exactly cancels out any and all vibrations from all directions?
My part is already done. You are the one tying to bring in mysterious variables here, in increasingly desperate tactics to claim that it is all too noisy to detect. You are the one who has no actual refutation, instead resting your argument on mysterious noise. Substantiate your mysterious variables.
I've shown you repeatedly where the noise comes from. The centripetal force generated by your own body when spun about an axis at 2+Hz, which is greater than 100x the acceleration difference we're supposed to be looking for. The sources/studies show we can detect 0.02 m/s^2 of vibrational acceleration. But that in no way shows we can tell the difference between 23.18 m/s^2 and 23.22 m/s^2, which would be the maximum percentage difference between front and back when spun at 2Hz. I've substantiated these variables in an earlier post and requested you to check my math. Or explain why you disagree with my necessary assumptions. You've done neither, suggesting you feel they are fine. Substantiate your claim that we can detect/feel that difference, or explain where I've erred in my math. Or I guess keep deliberately missing the point I suppose.
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There was already plenty of noise in the experiments in which it was determined that people can feel 0.01+ m/s^2. Bodily movements, attention, etc. Yet 0.01+ m/s^2 was still set as the threshold for human perception.
I checked, clothing is only mentioned in a test-safety sense in the human vibration book.
So, I will say my mysterious noise will be clothing. We all get one.
It's most relevant as it moves when you rotate. Literally making noise.
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My part is already done. You are the one tying to bring in mysterious variables here, in increasingly desperate tactics to claim that it is all too noisy to detect. You are the one who has no actual refutation, instead resting your argument on mysterious noise. Substantiate your mysterious variables.
I think Curious already touched on this a while back. But the data you're quoting from the "Handbook of Human Vibration" is from Chapter 6, “Perception of Whole-body Vibration and the Assessment of Vibration in Buildings". Is that really an appropriate comparison to earth rotation? Building vibration? Seems like an equally mysterious substantiation. Shouldn't you actually be quote mining from Chapter 7 "Motion Sickness", maybe specifically 7.4 "Effects of Rotational Motion"?
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But, and this is the thing you keep ignoring, MY MATHS IS WRONG. It’s wrong for the same reason the maths in the video is wrong. I have assumed that you are going at 460m/s in a STRAIGHT LINE in one direction and, over 12 hours you change velocity so that you are now going at 460m/s in the opposite direction. That is the logic of the video.
That's the error in the video, it only shows acceleration in "x-direction". What about the "y-direction"? On the 6am and 6pm positions, you would have to place the same arrows as on 12am and 12pm, but in perpendicular direction. This guy, who produced the video, suppressed this acceleration. If you now combine these vectors - decreasing with a cos-function, increasing with a sin-function - the resulting vector would be radial from (or to) earth center.
You could have it easier, if you consider the setup as this what it is: (sorry I don't know the exact term in English) a steady circular motion: The only force or acceleration produced by this motion is centripetal/centrifugal force:
a) radial to the center
b) constant
So at the equator the force is anti-parallel to gravity. The only thing you could measure - not notice - is that things at the equator are a bit less heavy, and this doesn't change in the course of a day (constant!). And most notable there's no additional acceleration in the direction along the curved path.
This is my understanding too, I believe the force because of the earth's rotation works in a direction opposite to gravity.
So you are 0.3% lighter at the equator than at the poles. Enough to measure, not enough to notice.
I don't believe there is any sideways force but I'm willing to be corrected if I've got that wrong.
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I've shown you repeatedly where the noise comes from. The centripetal force generated by your own body when spun about an axis a 2+Hz, which is greater than 100x the acceleration difference we're supposed to be looking for. The sources/studies show we can detect 0.02 m/s^2 of vibrational acceleration. But that in no way shows we can tell the difference between 23.18 m/s^2 and 23.22 m/s^2, which would be the minimum percentage difference between front and back when spun at 2Hz.
By this logic it should be impossible to tell the difference between 9.8 m/s^2 and 9.8 m/s^2 +/- 0.02 m/s^2. Is that correct?
Yet we previously saw that the first example used in the quora link was with sensing small vibrations in an elevator, which travels in a vertical direction.
We read the following:
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3189668/
Although studies have shown that the human perception threshold to vibration varies slightly with age, position of the body, body region, and axis of vibration,10,22 the median human vibration perception threshold is approximately 0.01 m/s^2 for vertical vibration
How is vertical vibration of that small magnitude detected if the body is already traveling vertically at 9.8 m/s^2?
This shows the premise, that additional accelerations cannot be felt, to be faulty.
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I've shown you repeatedly where the noise comes from. The centripetal force generated by your own body when spun about an axis a 2+Hz, which is greater than 100x the acceleration difference we're supposed to be looking for. The sources/studies show we can detect 0.02 m/s^2 of vibrational acceleration. But that in no way shows we can tell the difference between 23.18 m/s^2 and 23.22 m/s^2, which would be the minimum percentage difference between front and back when spun at 2Hz.
By this logic it should be impossible to tell the difference between 9.8 m/s^2 and 9.8 m/s^2 +/- 0.02 m/s^2. Is that correct?
Yet we previously saw that the first example used in the quora link was with sensing small vibrations in an elevator, which travels in a vertical direction.
We read the following:
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3189668/
Although studies have shown that the human perception threshold to vibration varies slightly with age, position of the body, body region, and axis of vibration,10,22 the median human vibration perception threshold is approximately 0.01 m/s^2 for vertical vibration
How is vertical vibration of that small magnitude detected if the body is already traveling vertically at 9.8 m/s^2?
This shows the premise, that additional accelerations cannot be felt, to be faulty.
This is a study about vibrations from nearby construction equipment. As well, humans are less affected than rats.
"Vibration caused by various items of construction equipment at 3 ft from the cage were evaluated relative to the RFR and SFR of humans, rats, and mice in 3 anatomic locations. In addition, the vibration levels in the RFR and SFR that resulted from the use of a large jackhammer and were measured at various locations and distances in the facility and evaluated in terms of humans, rats, and mice in 3 anatomic locations. Taken together, the data indicate that a given vibration source generates vibration in frequency ranges that are more likely to affect rats and mice as compared with humans."
As well, you left off a part of the sentence you quoted.
"...the median human vibration perception threshold is approximately 0.01 m/s2 for vertical vibration (the vibration measured in the current study) between 0 and 63 Hz."
As in the vibration from construction equipment. How is this relative to earth rotation?
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I've shown you repeatedly where the noise comes from. The centripetal force generated by your own body when spun about an axis a 2+Hz, which is greater than 100x the acceleration difference we're supposed to be looking for. The sources/studies show we can detect 0.02 m/s^2 of vibrational acceleration. But that in no way shows we can tell the difference between 23.18 m/s^2 and 23.22 m/s^2, which would be the minimum percentage difference between front and back when spun at 2Hz.
By this logic it should be impossible to tell the difference between 9.8 m/s^2 and 9.8 m/s^2 +/- 0.02 m/s^2. Is that correct?
Yet we previously saw that the first example used in the quora link was with sensing small vibrations in an elevator, which travels in a vertical direction.
We read the following:
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3189668/
Although studies have shown that the human perception threshold to vibration varies slightly with age, position of the body, body region, and axis of vibration,10,22 the median human vibration perception threshold is approximately 0.01 m/s^2 for vertical vibration
How is vertical vibration of that small magnitude detected if the body is already traveling vertically at 9.8 m/s^2?
This shows the premise, that additional accelerations cannot be felt, to be faulty.
This is a study about vibrations from nearby construction equipment. As well, humans are less affected than rats.
"Vibration caused by various items of construction equipment at 3 ft from the cage were evaluated relative to the RFR and SFR of humans, rats, and mice in 3 anatomic locations. In addition, the vibration levels in the RFR and SFR that resulted from the use of a large jackhammer and were measured at various locations and distances in the facility and evaluated in terms of humans, rats, and mice in 3 anatomic locations. Taken together, the data indicate that a given vibration source generates vibration in frequency ranges that are more likely to affect rats and mice as compared with humans."
As well, you left off a part of the sentence you quoted.
"...the median human vibration perception threshold is approximately 0.01 m/s2 for vertical vibration (the vibration measured in the current study) between 0 and 63 Hz."
As in the vibration from construction equipment. How is this relative to earth rotation?
The same perception threshold of 0.01 m/s^2 is given in the previous source I gave. (https://forum.tfes.org/index.php?topic=11079.msg169791#msg169791) The threshold is 0.01 m/s^2 is where people feel vibration. It doesn't matter if it comes from a jackhammer. That's the threshold.
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The same perception threshold of 0.01 m/s^2 is given in the previous source I gave. (https://forum.tfes.org/index.php?topic=11079.msg169791#msg169791) The threshold is 0.01 m/s^2 is where people feel vibration. It doesn't matter if it comes from a jackhammer. That's the threshold.
Is 'vibration' perception the same as "a spinning earth would add additional pull straight down towards its surface" perception?
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The same perception threshold of 0.01 m/s^2 is given in the previous source I gave. (https://forum.tfes.org/index.php?topic=11079.msg169791#msg169791) The threshold is 0.01 m/s^2 is where people feel vibration. It doesn't matter if it comes from a jackhammer. That's the threshold.
Is 'vibration' perception the same as "a spinning earth would add additional pull straight down towards its surface" perception?
The importance of vibration perception in this subject of sensing acceleration was described in the quora link (https://www.quora.com/What-is-the-minimum-amount-of-acceleration-that-a-human-can-detect).
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The same perception threshold of 0.01 m/s^2 is given in the previous source I gave. (https://forum.tfes.org/index.php?topic=11079.msg169791#msg169791) The threshold is 0.01 m/s^2 is where people feel vibration. It doesn't matter if it comes from a jackhammer. That's the threshold.
Is 'vibration' perception the same as "a spinning earth would add additional pull straight down towards its surface" perception?
The importance of vibration perception in this subject of sensing acceleration was described in the quora link (https://www.quora.com/What-is-the-minimum-amount-of-acceleration-that-a-human-can-detect).
Understood, but:
A) I wouldn't consider a Quora response to a question a substantiation of your position.
B) The Quora author 'speculates' and states: "Often it is the "jerk" that we notice in a situation like this (rather than the constant rate of acceleration itself)." When we talk about earth rotation, we are not talking about starting and stopping, but 'constant rate'.
C) 'vibration', at least to me, is not the same as a 'pull straight down towards its surface'. A vibration is, back and forth, up and down, side to side and everywhere in between. The other a pull, straight down to the surface. Seems like apples and oranges.
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The acceleration of an object moving in a circle can be determined by v2/r
https://www.physicsclassroom.com/class/circles/Lesson-1/Mathematics-of-Circular-Motion
v = 460m/s (velocity of earth's rotation at the equator)
r = 6371000 (radius of earth)
460x460/6371000 = 0.033213m/s/s
So that is the acceleration due to the rotation of the earth and that accelaration is directed inwards. That is counter-intuitive because you feel like you're being thrown outwards but
"The sensation of an outward force and an outward acceleration is a false sensation. There is no physical object capable of pushing you outwards. You are merely experiencing the tendency of your body to continue in its path tangent to the circular path along which the car is turning. You are once more left with the false feeling of being pushed in a direction that is opposite your acceleration."
https://www.physicsclassroom.com/class/circles/Lesson-1/The-Centripetal-Force-Requirement
So unless I've understood anything wrong, there is no sideways acceleration or force that you'd feel.
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I've shown you repeatedly where the noise comes from. The centripetal force generated by your own body when spun about an axis a 2+Hz, which is greater than 100x the acceleration difference we're supposed to be looking for. The sources/studies show we can detect 0.02 m/s^2 of vibrational acceleration. But that in no way shows we can tell the difference between 23.18 m/s^2 and 23.22 m/s^2, which would be the minimum percentage difference between front and back when spun at 2Hz.
By this logic it should be impossible to tell the difference between 9.8 m/s^2 and 9.8 m/s^2 +/- 0.02 m/s^2. Is that correct?
Yet we previously saw that the first example used in the quora link was with sensing small vibrations in an elevator, which travels in a vertical direction.
We read the following:
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3189668/
Although studies have shown that the human perception threshold to vibration varies slightly with age, position of the body, body region, and axis of vibration,10,22 the median human vibration perception threshold is approximately 0.01 m/s^2 for vertical vibration
How is vertical vibration of that small magnitude detected if the body is already traveling vertically at 9.8 m/s^2?
This shows the premise, that additional accelerations cannot be felt, to be faulty.
No, this shows there is a higher threshold than the 0.01-0.02 or so of the other study, when dealing with a difference in larger accelerations, rather than simply sensing it. The Quora link stated that sensing an elevator movement had a lower bounds of 0.1 m/s^2. Significantly higher than anything we've determined for what may or may not exist for the rotation of the Earth. I would argue this would be compounded depending on speed, but that's largely irrelevant to the fact his statements still fully support we should be unable to tell the difference between 9.8 m/s^2 and 9.8 m/s^2 +/- 0.02 m/s^2, or whatever speeds we might be discussing. We can sense vibrations as tiny as 0.01 m/s^2. That doesn't mean we can tell the difference between 0.03 and 0.04. Or anything else for that matter.
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The same perception threshold of 0.01 m/s^2 is given in the previous source I gave. (https://forum.tfes.org/index.php?topic=11079.msg169791#msg169791) The threshold is 0.01 m/s^2 is where people feel vibration. It doesn't matter if it comes from a jackhammer. That's the threshold.
Is 'vibration' perception the same as "a spinning earth would add additional pull straight down towards its surface" perception?
The importance of vibration perception in this subject of sensing acceleration was described in the quora link (https://www.quora.com/What-is-the-minimum-amount-of-acceleration-that-a-human-can-detect).
Understood, but:
A) I wouldn't consider a Quora response to a question a substantiation of your position.
B) The Quora author 'speculates' and states: "Often it is the "jerk" that we notice in a situation like this (rather than the constant rate of acceleration itself)." When we talk about earth rotation, we are not talking about starting and stopping, but 'constant rate'.
C) 'vibration', at least to me, is not the same as a 'pull straight down towards its surface'. A vibration is, back and forth, up and down, side to side and everywhere in between. The other a pull, straight down to the surface. Seems like apples and oranges.
Detecting a change in acceleration is a detection of vibration upon your body. Of course it is directly relevant. Seeing as the author of that article wrote a thesis on the matter, I would characterize the article as more than 'speculation'.
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The acceleration because of the rotation of the earth is vertical, not horizontal. There is no sideways pull on us because of it so far as I can gather from the links above about the physics of circular motion.
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it doesn't make any sense to compare a vibration to a constant centrifugal force.
a vibration is a change in acceleration. the centrifugal acceleration for a person standing on the earth's surface is constant. those are not comparable at all.
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The acceleration because of the rotation of the earth is vertical, not horizontal. There is no sideways pull on us because of it so far as I can gather from the links above about the physics of circular motion.
https://www.physicsclassroom.com/class/circles/Lesson-1/Acceleration
And since velocity is a vector that has both magnitude and direction, a change in either the magnitude or the direction constitutes a change in the velocity. For this reason, it can be safely concluded that an object moving in a circle at constant speed is indeed accelerating.
...
(http://www.physicsclassroom.com/Class/circles/u6l1b1.gif)
Direction of the Acceleration Vector
Note in the diagram above that there is a velocity change for an object moving in a circle with a constant speed. A careful inspection of the velocity change vector in the above diagram shows that it points down and to the left. At the midpoint along the arc connecting points A and B, the velocity change is directed towards point C - the center of the circle. The acceleration of the object is dependent upon this velocity change and is in the same direction as this velocity change. The acceleration of the object is in the same direction as the velocity change vector
it doesn't make any sense to compare a vibration to a constant centrifugal force.
a vibration is a change in acceleration. the centrifugal acceleration for a person standing on the earth's surface is constant. those are not comparable at all.
If you are standing with the left side of your body facing the acceleration and turn in a complete circle you would be experiencing a change of 200% of the acceleration on the left side of your body when you travel 180 degrees, and then another 200% as you return to your spot. The side of your body is being pulled and pushed in different directions.
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The acceleration because of the rotation of the earth is vertical, not horizontal. There is no sideways pull on us because of it so far as I can gather from the links above about the physics of circular motion.
https://www.physicsclassroom.com/class/circles/Lesson-1/Acceleration
And since velocity is a vector that has both magnitude and direction, a change in either the magnitude or the direction constitutes a change in the velocity. For this reason, it can be safely concluded that an object moving in a circle at constant speed is indeed accelerating.
...
(http://www.physicsclassroom.com/Class/circles/u6l1b1.gif)
Direction of the Acceleration Vector
Note in the diagram above that there is a velocity change for an object moving in a circle with a constant speed. A careful inspection of the velocity change vector in the above diagram shows that it points down and to the left. At the midpoint along the arc connecting points A and B, the velocity change is directed towards point C - the center of the circle. The acceleration of the object is dependent upon this velocity change and is in the same direction as this velocity change. The acceleration of the object is in the same direction as the velocity change vector
it doesn't make any sense to compare a vibration to a constant centrifugal force.
a vibration is a change in acceleration. the centrifugal acceleration for a person standing on the earth's surface is constant. those are not comparable at all.
If you are standing with the left side of your body facing the acceleration and turn in a complete circle you would be experiencing a change of 200% of the acceleration on the left side of your body when you travel 180 degrees, and then another 200% as you return to your spot. The side of your body is being pulled and pushed in different directions.
From the same article you cited:
"A careful examination of the flame reveals that the flame will point towards the center of the circle, thus indicating that not only is there an acceleration; but that there is an inward acceleration. This is one more piece of observable evidence that indicates that objects moving in a circle at a constant speed experience an acceleration that is directed towards the center of the circle."
Seems that acceleration is directed toward the center of the circle, not side to side, as Gary already pointed out above.
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Why would standing on a static planet create additional pull towards it's surface when the planet starts to rotate?
These articles are talking about the changing velocity angle being adjusted "towards the center" as the circle is put into motion, like with the two modes of v on the circle in the above illustration.
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Why would standing on a static planet create additional pull towards it's surface when the planet starts to rotate?
I believe these articles are talking about the changing velocity angle being adjusted "towards the center" as the circle is put into motion.
If I'm understanding the article correctly, it's talking about constant circular rotation, not when it starts to rotate. Which is what we are talking about; constant rotation. And I thought your point in citing the article was that circular rotation acceleration does not pull down to the center, but goes side to side. That's what I thought you meant by, "The side of your body is being pulled and pushed in different directions."
But that's not what the article is saying. It's saying, to Gary's point, it's pulling you toward the center.
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Please explain, in your own words, why putting a planet into a constant rotation would cause a greater pull directly downwards towards it's surface.
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Please explain, in your own words, why putting a planet into a constant rotation would cause a greater pull directly downwards towards it's surface.
What an interesting request.
Ok Professor, in my own words:
You start out with the premise that we should feel the rotation of earth in a horizontal manner. The example, putting 7 bucks in change in my pocket should make for the physical perception that if I move sideways in one direction or the other I should feel that weight, much like if I were standing upon a rotating earth. When the contention is that a constant rotating earth pulls you toward the center, not side to side, and has literally nothing to do with sideways/horizontal movement nor 7 dollars of change in my pocket.
Then you cite an article about how humans perceive the vibrations of a vibrating building. Which has nothing to do with a constant rotating earth.
Maths ensue, none of which on your part have to do with a constant rotating earth.
Then you cite a study that tested the vibration and its effects from nearby construction equipment on mice. Again, nothing to do with the constant rotation of an earth and it's effect on humans.
When questioned, you state, "Detecting a change in acceleration is a detection of vibration upon your body." Fair enough, but we've been talking all along about constant rotation, not a change. Again, your argument is not germane to the matter at hand; constant rotation.
Then you cite the https://www.physicsclassroom.com/class/circles/Lesson-1/Acceleration piece where you think it supports your contention that rotating is a horizontal vector velocity, when, in actuality, the article states that "objects moving in a circle at a constant speed experience an acceleration that is directed towards the center of the circle." This from the article you cited to seemingly bolster your position and it directly contradicts your position.
Then you go back to your static or "put in motion" notions when all along, for three pages, we've been talking about constant motion, constant rotation. Not starting, not stopping and not static, constant.
And lastly, if you feel I didn't already answer your question, "why putting a planet into a constant rotation would cause a greater pull directly downwards towards it's surface." The answer is: Gravity.
How is that Professor?
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Why would standing on a static planet create additional pull towards it's surface when the planet starts to rotate?
It is counter-intuitive, isn't it? And I had to think about it. I think the simple answer is there isn't an additional pull, there's just a pull because of gravity.
Newton's first law of motion states that a body will stay at rest or continue in a straight line unless acted on by a force.
So if you're travelling at 460m/s on the equator but you're on a revolving planet, and thus going round in a circle, then what is stopping you from continuing in a straight line and flying off into space?
There must be a force acting on you which keeps you stuck to the surface.
Luckily there is, it's called gravity.
If we were on a flat earth which had gravity pointing downwards uniformly - or if we were on a flat earth which was accelerating upwards - then there there would be no variation in g across that plane. Unless there are other forces at work and I know you have some fudge about "celestial gravitation" or something.
In the real world, in a rotating frame of reference the acceleration because of the rotation points inwards, thus the force points inwards. We feel that as an outward (not sideways) force which balances out the inwards force.
I guess you can think of it as most of the gravity is keeping you on the ground, but part of it is keeping you moving in a circle. That part is what we feel as centrifugal force
That's what makes your apparent weight lighter at the equator than at the poles. There is no sideways force. If you're on a merry go round you are thrown outwards, not sideways.
Some info here:
https://en.wikipedia.org/wiki/Centrifugal_force#Weight_of_an_object_at_the_poles_and_on_the_equator
And this video explains it quite well:
https://www.youtube.com/watch?v=Qwxl0EX-hDA
One important thing to note here - if you don't understand all this then that's fine. Honestly, I'm struggling to understand some of it myself.
But you not understanding it doesn't make it wrong.
A common FE argument is basically "I don't understand how..." and the conclusion is that the '...' most be wrong because the person doesn't understand it.
That's not how things work. The truth remains true regardless of what you believe or understand.
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The answer to this is not "it's just gravity."
Stack posted the following quote:
A careful examination of the flame reveals that the flame will point towards the center of the circle, thus indicating that not only is there an acceleration; but that there is an inward acceleration. This is one more piece of observable evidence that indicates that objects moving in a circle at a constant speed experience an acceleration that is directed towards the center of the circle.
How would objects moving in a circle at a constant speed experience an acceleration towards the center of the circle?
The answer is on the same page:
https://www.physicsclassroom.com/class/circles/Lesson-1/Acceleration
(https://i.imgur.com/1Ksfte1.png)
At the midpoint along the arc connecting points A and B, the velocity change is directed towards point C - the center of the circle.
It is not talking about a downwards pull towards the center of the circle at all. It's talking about an acceleration and velocity change that is "directed towards" the center as the circle rotates.
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The answer to this is not "it's just gravity."
Stack posted the following quote:
A careful examination of the flame reveals that the flame will point towards the center of the circle, thus indicating that not only is there an acceleration; but that there is an inward acceleration. This is one more piece of observable evidence that indicates that objects moving in a circle at a constant speed experience an acceleration that is directed towards the center of the circle.
How would objects moving in a circle at a constant speed experience an acceleration towards the center of the circle?
The answer is on the same page:
https://www.physicsclassroom.com/class/circles/Lesson-1/Acceleration
(https://i.imgur.com/1Ksfte1.png)
At the midpoint along the arc connecting points A and B, the velocity change is directed towards point C - the center of the circle.
It is not talking about a downwards pull towards the center of the circle at all. It's talking about an acceleration and velocity change that is "directed towards" the center as the circle rotates.
In the next lesson it states:
"And in accord with Newton's second law of motion, an object which experiences an acceleration must also be experiencing a net force. The direction of the net force is in the same direction as the acceleration. So for an object moving in a circle, there must be an inward force acting upon it in order to cause its inward acceleration."
https://www.physicsclassroom.com/Class/circles/u6l1c.cfm
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In the next lesson it states:
"And in accord with Newton's second law of motion, an object which experiences an acceleration must also be experiencing a net force. The direction of the net force is in the same direction as the acceleration. So for an object moving in a circle, there must be an inward force acting upon it in order to cause its inward acceleration."
https://www.physicsclassroom.com/Class/circles/u6l1c.cfm
And the very next three sentences.. "This is sometimes referred to as the centripetal force requirement. The word centripetal (not to be confused with the F-word centrifugal) means center seeking. For object's moving in circular motion, there is a net force acting towards the center which causes the object to seek the center."
Emphasis not mine. "Seeking" the center != additional straight downwards acceleration caused by circular motion. Velocities A and B in the illustration above are deflected towards the center by the circular motion.
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In the next lesson it states:
"And in accord with Newton's second law of motion, an object which experiences an acceleration must also be experiencing a net force. The direction of the net force is in the same direction as the acceleration. So for an object moving in a circle, there must be an inward force acting upon it in order to cause its inward acceleration."
https://www.physicsclassroom.com/Class/circles/u6l1c.cfm
And the very next three sentences.. "This is sometimes referred to as the centripetal force requirement. The word centripetal (not to be confused with the F-word centrifugal) means center seeking. For object's moving in circular motion, there is a net force acting towards the center which causes the object to seek the center."
Emphasis not mine. "Seeking" the center != additional straight downwards acceleration caused by circular motion. Velocities A and B in the illustration above are deflected towards the center by the circular motion.
Admittedly, I am lost. What is your contention here? Is it the word "pull" versus "seek"?
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In the next lesson it states:
"And in accord with Newton's second law of motion, an object which experiences an acceleration must also be experiencing a net force. The direction of the net force is in the same direction as the acceleration. So for an object moving in a circle, there must be an inward force acting upon it in order to cause its inward acceleration."
https://www.physicsclassroom.com/Class/circles/u6l1c.cfm
And the very next three sentences.. "This is sometimes referred to as the centripetal force requirement. The word centripetal (not to be confused with the F-word centrifugal) means center seeking. For object's moving in circular motion, there is a net force acting towards the center which causes the object to seek the center."
Emphasis not mine. "Seeking" the center != additional straight downwards acceleration caused by circular motion. Velocities A and B in the illustration above are deflected towards the center by the circular motion.
I'm also confused what you're arguing here. The part you've quoted - I've put a bit in bold - is saying exactly what I'm saying and what you appear to be arguing against. The force acts downwards, you feel that as a sensation as being thrown outwards, you don't get knocked sideways. If you've been on a roundabout you know that. And the point is the amount you feel that is around 0.3% of your weight. Enough to measure, not enough to feel.
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It looks like everybody here as an education on Newton's 2nd law. It really doesn't apply to the earth at all if, according to FET, the earth is not spinning. You should feel the vertical acceleration force at your feet and that's about it. It looks to me that Newton's 2nd law would be a critical factor for the moon under FET. Acceleration is a change in velocity. Velocity is also a vector so acceleration could be either a change in speed or a change in direction, or it could be both at the same time. Keeping the moon stationary over the flat earth requires a constant acceleration upwards (vertical relative to the surface of the earth). Keeping the moon in a circular path would require a force sideways (horizontal relative to the surface of the earth). What is the provider of that force? The force must be there because you can see a constant change in the moon's velocity, AKA a circular motion.