The Flat Earth Society

Flat Earth Discussion Boards => Flat Earth Media => Topic started by: Tom Bishop on October 19, 2018, 10:05:25 PM

Title: Heliocentric Speed Change Problem
Post by: Tom Bishop on October 19, 2018, 10:05:25 PM
An interesting video about how we should be able to feel the speed changes asserted by the heliocentric model.

https://www.youtube.com/watch?v=vxoSTRF92Y8

Title: Re: Heliocentric Speed Change Problem
Post by: RonJ on October 20, 2018, 04:52:55 AM
It's an interesting video.  I was taught in engineering school that force, velocity, and acceleration are all vectors.  As the earth rotates the acceleration slowly increases or decreases because the rotation velocity is being added to the earths orbital velocity.  However, its only the vector component that's with or against the vector of the earths orbital path that will count towards the acceleration you will feel.  I would expect to see a very slow increase in acceleration for 6 hours followed by a very slow decrease for the next 6 hours.  The force distributed over the surface area of your body would be like a gnat creeping on to you over a 6 hour period, then off again.  As long as the gnat didn't bite, I don't think that I would feel a thing.
Title: Re: Heliocentric Speed Change Problem
Post by: Tom Bishop on October 20, 2018, 06:06:41 PM
I converted the accelerations mentioned in the video to newtons and compared it to the weight of a quarter.

Acceleration at Equator = 21.5 mm / s^2 (from video)

Units of Force on 175 pound man via F=ma

(175 pound) * 21.5 (mm / (s^2)) = 1.70664129 newtons

1 US Quarter = 5.670 grams

US Quarter Weight in Newtons of Force

(5.67000 grams) * 9.8 (m / (s^2)) = 0.055566 newtons

1.70664129 / 0.055566 = 30.7138 quarters


Acceleration at 45 deg North = 10.75 mm / s^2 (from video)

Units of Force on 175 pound man via F=ma

(175 pound) * 10.75 (mm / (s^2)) = 0.853320646 newtons

1 US Quarter = 5.670 grams

US Quarter Weight in Newtons of Force

(5.67000 grams) * 9.8 (m / (s^2)) = 0.055566 newtons

0.853320646 / 0.055566 = 15.3569 quarters
Title: Re: Heliocentric Speed Change Problem
Post by: RonJ on October 20, 2018, 07:56:35 PM
My figures were close to what you got as well.  A McDonalds Big Mac has more mass and that doesn't even include the fries and shake.  It would have to be eaten over a 12 hour period.  Then after you got it down, you could then discharge about the same mass in the bathroom over the following 12 hours.  A human wouldn't feel a thing, except maybe in the belt line over a long period.  The percentage change of your total mass is very small. 
Title: Re: Heliocentric Speed Change Problem
Post by: HorstFue on October 21, 2018, 11:24:18 AM
Still falling down this pit hole?
Where are those 66,000MPH in the calculation for the acceleration?
Remove the Sun from the diagram, it makes no difference.
Then we are back at a spinning earth, rotating with a rim speed of about 1000MPH at the equator.
There are various threads related to this. For a given latitude this produces a constant centrifugal force, counteracting gravity, which is a tiny, measurable effect (hint: oblate spheroid).
Title: Re: Heliocentric Speed Change Problem
Post by: AllAroundTheWorld on October 21, 2018, 04:14:08 PM
Still falling down this pit hole?
Where are those 66,000MPH in the calculation for the acceleration?
Remove the Sun from the diagram, it makes no difference.
Then we are back at a spinning earth, rotating with a rim speed of about 1000MPH at the equator.
There are various threads related to this. For a given latitude this produces a constant centrifugal force, counteracting gravity, which is a tiny, measurable effect (hint: oblate spheroid).

This is a good example of FE videos which are of the form “the heliocentric model must be wrong because if it was right then ...”.

But the ... is always wrong and it’s wrong because of a poor understanding of physics. In this example if the 66,000 is constant then there’s no acceleration there, the only acceleration is the angular velocity from the rotation of the earth and since that is one revolution per 24 hours it is negligible.

EDIT:
If you’ve ever been on a plane which is stacking while in a queue to land the plane goes in circles. So using the above logic you’re going from about 500mph in one direction to 500mph in the other. So that’s a change of 1,000mph in what, 30 seconds. And yes, you feel the change of direction but not like you would if you were on a train going at 500mph and which could change direction in 30 seconds so it was going 500mph in the opposite direction. The radius of the bend changes the force you feel.

I’m going to start a thread about this, I think. Common FE straw men or fallacies.
Title: Re: Heliocentric Speed Change Problem
Post by: Tom Bishop on October 21, 2018, 08:03:22 PM
This kind of acceleration should be detectable on a stationary body, yet it doesn't even affect the leaning of a dandelion.
Title: Re: Heliocentric Speed Change Problem
Post by: AllAroundTheWorld on October 21, 2018, 08:27:58 PM
But the ... is always wrong and it’s wrong because of a poor understanding of physics.
QED, Tom. QED...
Title: Re: Heliocentric Speed Change Problem
Post by: Tom Bishop on October 21, 2018, 08:43:31 PM
But the ... is always wrong and it’s wrong because of a poor understanding of physics.
QED, Tom. QED...

Your last message"if the 66,000 is constant then there’s no acceleration there, the only acceleration is the angular velocity from the rotation of the earth" is already accounted for by the author. And "and since that is one revolution per 24 hours it is negligible" is false.

You have provided no demonstration that you have watched or understood the video, or demonstration of your conclusion. It appears that you have a long way to go for a career in physics. :(
Title: Re: Heliocentric Speed Change Problem
Post by: garygreen on October 21, 2018, 08:46:28 PM
the average acceleration due to gravity on the earth's surface is 9.8 m/s2.  adding 0.01 m/s2 is totally insignificant.

how insignificant?  using the numbers you provide, the difference in force on a person between 45 deg north and the equator is given by 1.707 N - 0.853 N = 0.854 N.  a weight of 0.854 N corresponds to a mass of 81 grams. 

81 grams is about 1.5 times the mass of a tennis ball.  let's round up to two.  and that small addition is distributed evenly throughout your whole body.  so in the time it takes to travel from ecuador to quebec, you'll feel two tennis balls heavier.  honestly, eating lunch is more noticeable than that.
Title: Re: Heliocentric Speed Change Problem
Post by: Tom Bishop on October 21, 2018, 08:59:00 PM
the average acceleration due to gravity on the earth's surface is 9.8 m/s2.  adding 0.01 m/s2 is totally insignificant.

how insignificant?  using the numbers you provide, the difference in force on a person between 45 deg north and the equator is given by 1.707 N - 0.853 N = 0.854 N.  a weight of 0.854 N corresponds to a mass of 81 grams. 

81 grams is about 1.5 times the mass of a tennis ball.  let's round up to two.  and that small addition is distributed evenly throughout your whole body.  so in the time it takes to travel from ecuador to quebec, you'll feel two tennis balls heavier.  honestly, eating lunch is more noticeable than that.

Except that in this case your 'lunch' isn't attempting to fall straight down with the rest of your body. If your stomach, or a distribution over your body, were attempting to fall to the horizontal, with the weight of 15 or 30 quarters, you would feel it. This phenomena should be detectable. There should be experiments which detect this.

This isn't a matter of "feeling a little heavier." What kind of characterization is that?


To the 'it is very gradual' comments:

If we were facing to the north and feeling pulled with the weight of 15 US quarters to our left, then we should be able to turn around to face the south and instantly feel pulled with the weight of 15 US quarters to our right. Yet we feel none of this.
Title: Re: Heliocentric Speed Change Problem
Post by: garygreen on October 21, 2018, 09:19:02 PM
Except that in this case your 'lunch' isn't attempting to fall straight down with the rest of your body.

what?  gravity also affects your lunch.  It appears that you have a long way to go for a career in physics. :(

This isn't a matter of "feeling a little heavier." What kind of characterization is that?

that's literally the definition of weight.  if you are accelerated downward more/less, then you weigh more/less.  you weigh more in quebec than you do in ecuador.
Title: Re: Heliocentric Speed Change Problem
Post by: Tom Bishop on October 21, 2018, 09:39:09 PM
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.
Title: Re: Heliocentric Speed Change Problem
Post by: AllAroundTheWorld on October 21, 2018, 09:58:15 PM
You have provided no demonstration that you have watched or understood the video
I don't know how I demonstrate that I watched the video.
I did though and I understood what he's getting at. His conclusion is false though because all he's doing is doing the maths as though the velocity changes in a straight line. It does not, it's a very slow circle. Slow in terms of angular velocity.
This does have an effect, but it's something like 0.3% of g, not something you would notice.
This is another things we can measure but not perceive.
Put your dandelion on a merry go round, now rotate that merry go round at one revolution per day and see how much effect that has.

There's some maths here which might help you

https://astronomy.stackexchange.com/questions/23489/why-doesnt-the-earths-rotation-throw-us-off-the-surface
Title: Re: Heliocentric Speed Change Problem
Post by: Tom Bishop on October 21, 2018, 10:22:53 PM
Put your dandelion on a merry go round, now rotate that merry go round at one revolution per day and see how much effect that has.

There's some maths here which might help you

https://astronomy.stackexchange.com/questions/23489/why-doesnt-the-earths-rotation-throw-us-off-the-surface

Consider the acceleration the dandelion would experience at the very center of the merry-go-round vs if the dandelion is placed further towards the edge. The dandelion will experience more acceleration the larger the merry-go-round is, and the Round Earth is much larger than a merry-go-round. That the merry-go-round rotates at "one revolution per day" is an attempted trivialization of the matter and avoidance of the situation.

You should do the math on this, preferably in the appropriate forum.
Title: Re: Heliocentric Speed Change Problem
Post by: RonJ on October 21, 2018, 10:43:03 PM
As much as I hate to bring this up on this web site, there is one instrument that can detect the spinning of the earth.  It's been around for a long time, and can still be seen in museums.  Foucault's pendulum can detect and show that the earth is in rotation.  I'm looking at the physics of this now.  From my first quick look I can say that the case is clear, there was a pendulum built at the Amundesen-Scott South Pole station by some scientists and the results were as expected. The earth rotates in the expected direction and the acceleration of gravity is about as expected as well.  Effectively a pendulum is like a gyroscope, but just intermittently.  The angular momentum vector changes direction periodically as the mass at the end of the wire reverses direction.  Most of the pendulums built are at lower latitudes and behave a bit differently as expected.  What I don't know, yet, is the effect of the pendulum on the flat earth model.  I haven't used my Mathcad program in a while, but I'm hoping to construct some simulations that will give me an idea of the expected differences between the round earth, and flat earth models.     
Title: Re: Heliocentric Speed Change Problem
Post by: Tom Bishop on October 21, 2018, 11:19:31 PM
Here is our Foucault Pendulum page: https://wiki.tfes.org/Foucault_Pendulum

Feel free to address it in the appropriate forum.
Title: Re: Heliocentric Speed Change Problem
Post by: inquisitive on October 21, 2018, 11:38:09 PM
Time for the US to change to metric units like almost all of the rest of the world, a simpler system.
Title: Re: Heliocentric Speed Change Problem
Post by: AllAroundTheWorld on October 22, 2018, 08:25:16 AM
Consider the acceleration the dandelion would experience at the very center of the merry-go-round vs if the dandelion is placed further towards the edge. The dandelion will experience more acceleration the larger the merry-go-round is, and the Round Earth is much larger than a merry-go-round. That the merry-go-round rotates at "one revolution per day" is an attempted trivialization of the matter and avoidance of the situation.
Well, no, it's just explaining how physics actually works rather than how you are imagining it works.

And it's interesting you're trying to patronise me when you've gone and used pounds in F = ma above, not understanding that the unit of mass in that equation is kilograms :)

But anyway, here's my maths. At the equator the earth spins at 460m/s
So that's a 920m/s difference over about 12 hours.
There are 86400 seconds in 24 hours.
86400 / 2 = 43200s

So the accelaration is: 920/43200 = 0.0212m/s/s


F = ma

I'm around 69kgs so

69 x 0.0212 =  1.46N

And a Newton is "the force that an average sized apple makes on your hand when you hold it".

https://www.quora.com/How-much-force-is-1-Newton

As gary points out elsewhere, that's not just focused on your hand either but throughout your body.

With cars they go on about how fast a car can go from 0-60mph.
60mph is 26.82m/s so at the acceleration rate we're talking about above you'd go from 0-60 in about 21 minutes. I'm not sure that's something you'd easily feel.

Also, there is a problem in all the above. The problem is all the above assumes that the acceleration is in a straight line.
It isn't. That's the whole point, the acceleration occurs because of the earth's rotation so you have to use different maths which takes that into account.

The proper maths is here:

https://www-spof.gsfc.nasa.gov/stargaze/Srotfram1.htm

"Comparing this to the acceleration of gravity--say 9.81 m/s2--it is only 0.00346 or 0.346%. Effective gravity on the equator is reduced by the rotation, but only by about 1/3 of a percent"

So yes, it has an effect, but not one you can discern.
Title: Re: Heliocentric Speed Change Problem
Post by: Tom Bishop on October 22, 2018, 05:36:31 PM
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.
Title: Re: Heliocentric Speed Change Problem
Post by: shootingstar on January 04, 2019, 02:30:53 PM
You opened this thread with a video link Tom so here's another one with a different view to you. 

https://www.youtube.com/watch?v=_sbVNOyStqs



There are subtle clues visible in the sky over the course of a year that show us that the Earth is moving through space at said speeds.  No equipment other than your own eyes needed to see them.
Title: Re: Heliocentric Speed Change Problem
Post by: spherical on May 03, 2019, 09:12:38 PM
Years ago there was a student that came up with a intriguing question. He said: 

A car is running at 60km/h in a perfect straight concrete road, all four tires are touching the ground.
The bottom of the tires that touches the ground are unquestionable stopped in reference to the ground.
The upper part of the tires are speeding at twice the speed of the car.
The front of each tire, speeds as the car, but reducing speed, they need to stop when touching the ground.
The back of each tire, speeds as the car, but increasing speed, they need to be double speed on top.
The question was:  how the tire stands this constant acceleration and breaking once per revolution, without damaging? and why we don't feel such "bumping" speed change in the tires?

It took a while to make the students to "see the magic". 
Our brain is unable to by itself catch the idea of certain physical forms or movements.
The only way to start to grasp such light, is thinking; after the car/airplace accelerated and reach some steady speed, it flew few inches in the air.
During this minute, even that the tires are continuing rotating and "almost touching" the ground, and coincidentally they seems to be touching the ground and stopped while doing that, they are not.
In reference to the car and shaft, the tires are in a perfect circular motion, no breaking, no accelerating whatsoever. The centrifugal force on the rubber is perfectly balanced.  What is breaking is the reference to the ground, and as a reference it does not, and can not change anything in the rubber movement.  The surface and angular speed of any part of the tire rubber is exactly the same in reference to the shaft and car.

(https://media.wired.com/photos/5926cad8cefba457b079b25e/master/pass/GettyImages-536814811.jpg)

The same, the illusion of Earth's different speed on the illusion of forward and backward motion due its spinning.  You can find a ton of formulas and calculate several variables and came up to nice conclusions, but in real, it doesn't happen, you can't even measure it, because it doesn't exist.  It doesn't change a bit if the Earths is moving forward or backward in its path around the Sun, the spinning around its own axis cancels any angular vector.  The centrifugal force is equal based on the frame of reference, and doesn't change if you have the pivoting point moving in a constant speed.  What you have is a constant centrifugal force, that in case of Earth's is minuscule compared to the space deformation sliding vector.  Of course, you can have differences IF the pivoting point speed changes, that is not the case of Earth.

You can make a nice and easy experiment.  Remote the front bicycle wheel from the bicycle, cut 20 pieces of a light string 50 cm each, find 20 objects like large metallic nuts or solid 50+g objects (same size rocks or even little bags of sand), tie the objects to the end of the strings, and tie the other side of each string to the bicycle wheel equally spaced by 18 degrees.   Now, hold the wheel horizontally and rotate it until all the strings and objects spread aside in an angle around 45° by the centrifugal force, so you will be able to see any difference of speed, if they speed up the string will climb to a higher angle, if they slow down, the string and the object will go down.  Now, while it rotates happily, and holding the wheel in a way you could walk, start to walk forward holding the rotating wheel, keep an eye on the angle of each string.  Adjust your walking speed in order for the "backward motion" coincide with the ground, so they appear as stopped, and tell me if any string performing the "backward" movement goes down due low speeding or the forward movement would climb in angle for having a greater speed.  They would not.  If you run much fast, or attach the wheel to the hood of your car and run, the air movement will interfere with the results. 
Title: Re: Heliocentric Speed Change Problem
Post by: QED on May 04, 2019, 03:01:19 PM
This is by a wide margin the most accurate and complete explanation for relative motion I have read on this forum.

You certainly must have taught physics or have equivalent training. This makes it even more strange to me your prior diatribe regarding gravitation. It was filled with imprecise language, poorly constructed arguments, and an apparent ignorance of the normal force - which one learns in physics I.

Can you help reconcile for me this difference? Are there two of you?

In any event, I redact my stipulation to leave the physics to me. You clearly have valuable (albeit inconsistent) input.
Title: Re: Heliocentric Speed Change Problem
Post by: spherical on May 04, 2019, 04:34:41 PM
Thank you, QED.   I wish we could have only good days in life.