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Offline Tumeni

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #120 on: April 23, 2018, 10:29:12 PM »
Again, I see that you guys are trying to divide smaller numbers by bigger numbers to disprove this. You can't do that in division.

Why not? Says who, apart from you?

Are you telling me I can't divide 1 by 4 to yield 0.25, or one quarter anymore?


1 Year = 365.24 Days
1 Days = 24 Hours

365.24 Days / 24 Hours <--- You say this is invalid

It means nothing to everyone else. What does it mean to you? 365.24/24 = 1/24th of 365.24. Nothing else.

1 Dollar = 10 Dimes
1 Dime = 10 Pennies
10 Dimes / 10 Pennies = 1  <--- But this is correct. 10 Dimes fits into 10 Pennies 1 time

Once again,

1 Pound = 20 shillings
1 Shilling = 12 Pence

20 / 12 = 1.66666 - Your example only comes out with a whole number because of the 10/10 relationship of cents to dimes AND dimes to dollars



1 Yard = 5280 Feet
1 Foot = 12 inches

5280 Feet / 12 inches = 440  <--- This is correct as well. 12 Inches can fit neatly into 5280 feet 440 times

Total hogwash. 1 yard is 3 feet. 440 yards is a quarter mile, 5280 feet is 1 mile

This one only works out because

1 foot = 12 inches
3 feet = 1 yard
440 yards = quarter mile
1760 yards = 1 mile, so (1760*3) 5280 feet = 1 mile

Divide 5280 feet by 12 and you get 440 feet. Which is the same result as dividing 5280 by 4 to get one quarter mile (= 440 yards), then dividing that by 3 to get 440 feet (4*3 = 12).

You've picked another where the relationship between the units is built upon the figure that you're dividing into them, so it will naturally work out as a whole number.

Try it on anything where the relationship is not built this way, and it falls apart. See my example above.
« Last Edit: April 23, 2018, 10:45:20 PM by Tumeni »
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Offline Tom Bishop

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #121 on: April 23, 2018, 10:38:41 PM »
My mistake. You are correct. I meant to write "12 fits neatly into 5280 440 times"

----

Quote
1 Pound = 20 shillings
1 Shilling = 12 Pence

20 / 12 = 1.66666

This is actually correct. If you have 12 groups of 12 units, it doesn't fit into 20 x 12. The labels are arbitrary and do not represent real world things of real units.
« Last Edit: April 23, 2018, 10:49:12 PM by Tom Bishop »

Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #122 on: April 23, 2018, 10:40:52 PM »
Again, I see that you guys are trying to divide smaller numbers by bigger numbers to disprove this. You can't do that in division. The first number needs to be the biggest hierarchical group and the largest number.


You need to multiply the day by 24 as well to maintain the equivalence.

Tom there is NO equivalence! You're trying to balance two completely separate events! If this thread was an animal any decent vet would put it out of its misery... :-[


1 Year = 365.24 Days
1 Days = 24 Hours

365.24 Days / 24 Hours <--- You say this is invalid

But look at these:

--- --- ---

1 Dollar = 10 Dimes
1 Dime = 10 Pennies

10 Dimes / 10 Pennies = 1  <--- But this is correct. We got a whole number. 10 Dimes fits into 10 Pennies 1 time.

--- --- ---

1 Yard = 5280 Feet
1 Foot = 12 inches

5280 Feet / 12 inches = 440  <--- This is correct as well. We got a whole number. 12 inches can fit neatly into 5280 feet 440 times.

The 1 Foot = 12 inches is implicit in the above equation.

--- --- ---

I just divided unlike units and got a right answer  :o
I just want to point out this dimes/pennies one. Because your answer is total hogwash, and I can't believe you don't see it. 10 dimes doesn't fit into 10 pennies 1 time. 10 pennies doesn't fit into 10 dimes 1 time either. You've provided the exact example to help show your final statement false, within your attempt to prove it true. :O indeed. Not to mention none of these have any sort of units attached to them.

I like your last one too. Feet/inches. Just what is your unit here? 12 fits nicely into 5280, and? You've created a nonsense number again. It doesn't actually have any meaning beyond the number of times 12 fits into 5280. It's not the number of inches in 5280 feet, its not the number of feet in 5280 feet. It's a relatively meaningless number.

My mistake. I meant to write "12 fits neatly into 5280 440 times"

----

Quote
1 Pound = 20 shillings
1 Shilling = 12 Pence

20 / 12 = 1.66666

This is actually correct. If you have 12 groups of 12 units, it doesn't fit into 20. The values are arbitrary and do not represent real world things of real units.
Oh Tom, this is the best yet! Pound, pence, and shilling aren't real world things of real units? According to who? Do they not exist in your world? They're just a 'fake' as dimes, pennies, and dollars!

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Offline Tom Bishop

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #123 on: April 23, 2018, 10:43:58 PM »
This applies to measuring systems that are constant as ratios.

The dollar is divided by dimes and pennies, so it will work.

The mile is divided by foot and by inches, so it will work.

British currency values are not based on constants like that.
« Last Edit: April 23, 2018, 10:55:25 PM by Tom Bishop »

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Offline Tumeni

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #124 on: April 23, 2018, 10:51:29 PM »
This applies to measuring systems that are constant.

The dollar is divided by dimes and pennies, so it will work.

The mile is divided by foot and by inches, so it will work.

British currency values are not based on constants like that.

The pound was divided by shillings and pennies, in the same way that your dollar is divided by dimes and cents/pennies.

However;

1 Dollar = 4 Quarters
1 Quarter = 25 Cents

So ....  4/25 = ... what?
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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #125 on: April 23, 2018, 10:55:21 PM »
This applies to measuring systems that are constant.

The dollar is divided by dimes and pennies, so it will work.

The mile is divided by foot and by inches, so it will work.

British currency isn't based on constants like that.
Ah, so if it doesn't support your arbitrary hypothesis, we can just toss it out. Got it.

I guess we're also going to ignore how none of your answers are 'right' in so far as actually describing any sort of unit then, as I also pointed out? That they're just largely meaningless numbers?

Or maybe we should discuss the fact that the reason for your dismissal of British currency applies equally to rotational duration and orbital duration?

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Offline nickrulercreator

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #126 on: April 23, 2018, 10:59:46 PM »
Can someone give me a TLDR of the thread? I'm really not getting the argument that Tom is trying to make.
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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #127 on: April 23, 2018, 11:01:54 PM »
tom's argument (correct me if i'm wrong, tom) is that there must be an integer number of rotations in a single orbital period.

i still don't understand where he's getting that from.
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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #128 on: April 23, 2018, 11:03:51 PM »
Can someone give me a TLDR of the thread? I'm really not getting the argument that Tom is trying to make.
TL;DR (I think) Because the number of hours in a day don't go neatly into the number of days in a year, the RE model can't be correct.

If I've erred, a correction would be greatly appreciated however.

tom's argument (correct me if i'm wrong, tom) is that there must be an integer number of rotations in a single orbital period.

i still don't understand where he's getting that from.
I think it has to do with solar noon year to year. He doesn't seem to get that solar noon on important days (like the equinoxes) don't happen at the same time on the calendar every year. That, in fact, the only reason they even stay close is because of adjustments such as leap year.

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Offline nickrulercreator

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #129 on: April 23, 2018, 11:07:33 PM »
tom's argument (correct me if i'm wrong, tom) is that there must be an integer number of rotations in a single orbital period.

i still don't understand where he's getting that from.

I don't either. Why does he think this should be if he thinks this? Does he not know leap years exist to act as a correction?
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Offline Tom Bishop

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #130 on: April 24, 2018, 12:19:49 AM »
This applies to measuring systems that are constant.

The dollar is divided by dimes and pennies, so it will work.

The mile is divided by foot and by inches, so it will work.

British currency isn't based on constants like that.
Ah, so if it doesn't support your arbitrary hypothesis, we can just toss it out. Got it.

I guess we're also going to ignore how none of your answers are 'right' in so far as actually describing any sort of unit then, as I also pointed out? That they're just largely meaningless numbers?

Or maybe we should discuss the fact that the reason for your dismissal of British currency applies equally to rotational duration and orbital duration?

The rotational and orbital duration should have constant ratios. It is not merely arbitrary that the number of hours in a day should fit into the number of days in a year.

This applies to measuring systems that are constant.

The dollar is divided by dimes and pennies, so it will work.

The mile is divided by foot and by inches, so it will work.

British currency values are not based on constants like that.

The pound was divided by shillings and pennies, in the same way that your dollar is divided by dimes and cents/pennies.

However;

1 Dollar = 4 Quarters
1 Quarter = 25 Cents

So ....  4/25 = ... what?

As I described earlier, the first number needs to be the larger one and the greatest hierarchical unit in the scenario.


Can someone give me a TLDR of the thread? I'm really not getting the argument that Tom is trying to make.
TL;DR (I think) Because the number of hours in a day don't go neatly into the number of days in a year, the RE model can't be correct.

If I've erred, a correction would be greatly appreciated however.

tom's argument (correct me if i'm wrong, tom) is that there must be an integer number of rotations in a single orbital period.

i still don't understand where he's getting that from.
I think it has to do with solar noon year to year. He doesn't seem to get that solar noon on important days (like the equinoxes) don't happen at the same time on the calendar every year. That, in fact, the only reason they even stay close is because of adjustments such as leap year.

The Solar Year says that the sun should be in the same place in the sky after 1 Solar Year.


A visualization.

The sun travels across the earth's surface each day.

Earth circumference = 24,901 mi. In 1 Day the sun travels over 24,901 mi. of earth.

24,901 / 24 = 1037.54166667 miles. Over 1 hour the sun travels over 1037.54166667 miles

After 365 days: 24,901 mi. x 365 days = 9088865 miles

After 365.24  days: 24901 x 365.24 = 9094841.24 miles

Difference = 5976.24 miles

5976.24 miles / 1037.54166667 miles = 5.76. The hours in miles fits into the difference by 5.76 times. Where are those extra hours coming from? The sun will not be in the same place.
« Last Edit: April 24, 2018, 04:03:44 AM by Tom Bishop »

Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #131 on: April 24, 2018, 12:43:49 AM »
The Solar Year says that the sun should be in the same place in the sky after 1 year.

this is the discrepancy.  that's not really true.  you're conflating two different quantities.

what i mean to say is: the solar year is not precisely one period long.  the sun is not in exactly the same place in the sky after one period of the earth's orbit.  there is not an integer number of rotations in one period.

this article does a pretty good job of describing all of the different ways to define a year: http://www.orlandosentinel.com/news/space/go-for-launch/os-think-you-know-how-many-days-are-in-a-year-think-again-20170228-story.html

none of them are an integer numbers of days.  that's not part of the model.

« Last Edit: April 24, 2018, 12:46:09 AM by garygreen »
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Offline Tom Bishop

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #132 on: April 24, 2018, 12:46:21 AM »
The Solar Year says that the sun should be in the same place in the sky after 1 year.

this is the discrepancy.  that's not really true.  you're conflating two different quantities.

what i mean to say is: the solar year is not precisely one period long.  the sun is not in exactly the same place in the sky after one period of the earth's orbit.  there is not an integer number of rotations in one period.

What do you mean? The sun isn't in the same place in the sky after a Solar Year? It says that in the definition:

https://en.wikipedia.org/wiki/Tropical_year

Quote
A tropical year (also known as a solar year) is the time that the Sun takes to return to the same position in the cycle of seasons, as seen from Earth; for example, the time from vernal equinox to vernal equinox, or from summer solstice to summer solstice.]A tropical year (also known as a solar year) is the time that the Sun takes to return to the same position in the cycle of seasons, as seen from Earth; for example, the time from vernal equinox to vernal equinox, or from summer solstice to summer solstice.

Quote from: garygreen
this article does a pretty good job of describing all of the different ways to define a year: http://www.orlandosentinel.com/news/space/go-for-launch/os-think-you-know-how-many-days-are-in-a-year-think-again-20170228-story.html

none of them are an integer numbers of days.  that's not part of the model.

The Solar Day should fit into the Solar Year. The Solar Day is 24 hours long.
« Last Edit: April 24, 2018, 12:52:22 AM by Tom Bishop »

Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #133 on: April 24, 2018, 12:58:53 AM »
What do you mean? The sun isn't in the same place in the sky after a Solar Year? It says that in the definition:

https://en.wikipedia.org/wiki/Tropical_year

yes, that's the definition of a tropical year; but, i mean that one tropical year is not precisely the same duration as the time it takes the earth to go around the earth one time.

one tropical year is not equal to one period (one orbit of the earth around the sun).

the earth does not rotate on its axis an integer number of times in one orbit of the sun.
« Last Edit: April 24, 2018, 01:00:26 AM by garygreen »
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Offline Tom Bishop

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #134 on: April 24, 2018, 01:18:35 AM »
What do you mean? The sun isn't in the same place in the sky after a Solar Year? It says that in the definition:

https://en.wikipedia.org/wiki/Tropical_year

yes, that's the definition of a tropical year; but, i mean that one tropical year is not precisely the same duration as the time it takes the earth to go around the earth one time.

one tropical year is not equal to one period (one orbit of the earth around the sun).

the earth does not rotate on its axis an integer number of times in one orbit of the sun.

It says right in the link that a Solar Year is 354.24 days. Why isn't it equal to one sun rotation around the earth if the sun needs to get back into the same place in the sky according to the definition?

Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #135 on: April 24, 2018, 01:30:16 AM »
It says right in the link that a Solar Year is 354.24 days. Why isn't it equal to one sun rotation around the earth if the sun needs to get back into the same place in the sky according to the definition?

because the earth moves a little bit along its orbit during one rotation.  i think someone already posted this, but here is it again:


also, a solar day is not equal to one rotation around the earth's axis.  that's a sidereal day.

the earth does not rotate on its axis an integer number of times in one orbit, so neither of these (solar or sidereal days) are evenly divisible into one period.
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Offline Tom Bishop

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #136 on: April 24, 2018, 01:37:15 AM »
It says right in the link that a Solar Year is 354.24 days. Why isn't it equal to one sun rotation around the earth if the sun needs to get back into the same place in the sky according to the definition?

because the earth moves a little bit along its orbit during one rotation.  i think someone already posted this, but here is it again:


I don't see how it matters if the radius of the circle of the earth's path is 1 million miles or 1 mile. A year has 364.24 days.

The orbital path is elongated, however. This is true.

On Page 3 I posted a quote showing how little the Solar Year varies from where you measure it:

Quote from: Tom Bishop
And the Topical Year varies on solar return points (which is why they called it the 'mean' tropical year):

http://calendars.wikia.com/wiki/Tropical_year
Quote
    Current values and their annual change of the time of return to the cardinal ecliptic points[2] are:

        vernal equinox: 365.24237404 + 0.00000010338×a days
        northern solstice: 365.24162603 + 0.00000000650×a days
        autumn equinox: 365.24201767 − 0.00000023150×a days
        southern solstice: 365.24274049 − 0.00000012446×a days

It varies very little.

Quote
also, a solar day is not equal to one rotation around the earth's axis.  that's a sidereal day.

the earth does not rotate on its axis an integer number of times in one orbit, so neither of these (solar or sidereal days) are evenly divisible into one period.

The Sidrael Day is a rotation calculated based how much the stars move in relation to the sun. The stars move slightly slower than the sun. Lets assume that the stars don't exist and only focus on Solar Day. According to Solar Day the sun moves around the earth once in 24 hours. Form the last exercise I did with miles around the earth's circumference, which showed the sun traveling over all of the earth's circumference, we can see that the sun is offset at the end.

The sun should have returned to the same position over the earth. It did not.
« Last Edit: April 24, 2018, 01:50:41 AM by Tom Bishop »

Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #137 on: April 24, 2018, 01:55:36 AM »
I don't see how it matters if the radius of the circle of the earth's path is 1 million miles or 1 mile. A year has 364.24 days.

you're missing the point.  because the earth moves along its orbit (whatever the size) over the course of a rotation, a solar day is necessarily longer than a single rotation.

we can see that the sun is offset at the end.

yes, it is.  the earth does not rotate on its axis an integer number of times in one orbit of the sun.  imagine a planet that rotates 1.5 times over one orbit of its sun.  at the end of one orbit, the sun will be offset from its starting point.

The sun should have returned to the same position over the earth. It did not.

there is no such requirement in ret.
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Offline Tom Bishop

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #138 on: April 24, 2018, 02:12:17 AM »
I don't see how it matters if the radius of the circle of the earth's path is 1 million miles or 1 mile. A year has 364.24 days.

you're missing the point.  because the earth moves along its orbit (whatever the size) over the course of a rotation, a solar day is necessarily longer than a single rotation.

Quote
we can see that the sun is offset at the end.

yes, it is.  the earth does not rotate on its axis an integer number of times in one orbit of the sun.  imagine a planet that rotates 1.5 times over one orbit of its sun.  at the end of one orbit, the sun will be offset from its starting point.

I am talking about Solar Days. The Earth rotates in one rotation, and the sun makes one rotation around the earth, over a Solar Day.

The sun should have returned to the same position over the earth. It did not.

there is no such requirement in ret.

That's the definition of a Solar Year. The time it takes the sun to return to its same position. I have given you the quote.

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Offline Tom Bishop

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #139 on: April 24, 2018, 02:14:50 AM »
I don't see how it matters if the radius of the circle of the earth's path is 1 million miles or 1 mile. A year has 364.24 days.

you're missing the point.  because the earth moves along its orbit (whatever the size) over the course of a rotation, a solar day is necessarily longer than a single rotation.

It says that the Solar Year is 364.24 Solar Days. Now you are telling me that a Solar Year is a different number of Solar Days?

Quote
we can see that the sun is offset at the end.

yes, it is.  the earth does not rotate on its axis an integer number of times in one orbit of the sun.  imagine a planet that rotates 1.5 times over one orbit of its sun.  at the end of one orbit, the sun will be offset from its starting point.

I am talking about Solar Days. The Earth rotates in one rotation in relation to the Sun, and the Sun makes one rotation around the earth, over a Solar Day.

Quote
The sun should have returned to the same position over the earth. It did not.

there is no such requirement in ret.

Yes, it is the definition of a Solar Year. It is defined as the time it takes the sun to return to its same position. I have given you the quote.
« Last Edit: April 24, 2018, 02:18:43 AM by Tom Bishop »