What is the Linear speed of Sun in FE?
« on: September 28, 2017, 05:36:29 AM »
Based on below calculation what is the Linear speed of Sun in FE? If FEer don't use below method, what they use and can that be described here?

Linear Speed: The linear speed of a point on a rotating object depends on its distance from the center of rotation. The angular speed is the angle that an object moves through in a certain amount of time. The angular speed has units of radians per second (rad/s). There are 2π radians in a full circle. At a distance r from the center of the rotation, a point on the object has a linear speed equal to the angular speed multiplied by the distance r. The units of linear speed are meters per second, m/s.

linear speed = angular speed x radius of the rotation

v = ωr

v = linear speed (m/s)

ω = angular speed (radians/s)

r = radius of the rotation (m)

Re: What is the Linear speed of Sun in FE?
« Reply #1 on: September 28, 2017, 12:46:01 PM »
Based on below calculation what is the Linear speed of Sun in FE? If FEer don't use below method, what they use and can that be described here?

Linear Speed: The linear speed of a point on a rotating object depends on its distance from the center of rotation. The angular speed is the angle that an object moves through in a certain amount of time. The angular speed has units of radians per second (rad/s). There are 2π radians in a full circle. At a distance r from the center of the rotation, a point on the object has a linear speed equal to the angular speed multiplied by the distance r. The units of linear speed are meters per second, m/s.

linear speed = angular speed x radius of the rotation

v = ωr

v = linear speed (m/s)

ω = angular speed (radians/s)

r = radius of the rotation (m)
Based on the 'traditional' unipolar model it would be roughly 52 m/s. The radius is 12000 miles, and it makes one rotataion per 24 hours. Now mind you that's more of an average. It will move faster during the winter in the Northern hemiplane, and slower during the Northern hemiplane's summer. The bipolar model would be roughly the same 52 m/s, but it wouldn't ever get faster, only slower. So that would be more like the upper bounds.

Keep in mind I'm not sure how many of these numbers are FES 'valid' as they seem to have some disagreement on distance that has never been worked out. But the 12000 mile radius is what's used in their sun height calculation on the wiki.

Offline 3DGeek

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Re: What is the Linear speed of Sun in FE?
« Reply #2 on: September 28, 2017, 02:08:21 PM »
Based on below calculation what is the Linear speed of Sun in FE? If FEer don't use below method, what they use and can that be described here?

Linear Speed: The linear speed of a point on a rotating object depends on its distance from the center of rotation. The angular speed is the angle that an object moves through in a certain amount of time. The angular speed has units of radians per second (rad/s). There are 2π radians in a full circle. At a distance r from the center of the rotation, a point on the object has a linear speed equal to the angular speed multiplied by the distance r. The units of linear speed are meters per second, m/s.

linear speed = angular speed x radius of the rotation

v = ωr

v = linear speed (m/s)

ω = angular speed (radians/s)

r = radius of the rotation (m)
Based on the 'traditional' unipolar model it would be roughly 52 m/s. The radius is 12000 miles, and it makes one rotataion per 24 hours. Now mind you that's more of an average. It will move faster during the winter in the Northern hemiplane, and slower during the Northern hemiplane's summer. The bipolar model would be roughly the same 52 m/s, but it wouldn't ever get faster, only slower. So that would be more like the upper bounds.

Keep in mind I'm not sure how many of these numbers are FES 'valid' as they seem to have some disagreement on distance that has never been worked out. But the 12000 mile radius is what's used in their sun height calculation on the wiki.

There is some stuff about this on the Wiki: https://wiki.tfes.org/Constant_Speed_of_the_Sun

There are some epic quotes there: "The sun moves constant speed into the horizon at sunset because it is at such a height that already beyond the apex of perspective lines. It has maximized the possible broadness of the lines of perspective in relation to the earth. It is intersecting the earth at a very broad angle.".

Ohhh....OK...that explains it all PERFECTLY!   (**NOT**)

Then there are a bunch of total LIES about the way airplanes seem to cross the sky.  Clearly whoever wrote it never used a protractor and a stopwatch to check their claim...as usual, they are just guessing.

Again, we have a diagram with a vanishing point that is not at infinity...and a bunch of random words that actually CONTRADICT the claims they've just been making.

BOTTOM LINE:  Magic perspective makes everything work out perfectly, try not to worry your pretty little heads about it.  Tom understands it so it must be OK.

Hey Tom?  Still ducking my question about the path the photons take from the sun to my eye at sunset?   Yeah - I thought so.

(Oooohhhh - now there's a thought - if we add this question to the bottom of every post - do you think he'll stop responding to ALL of our threads?)
Hey Tom:  What path do the photons take from the physical location of the sun to my eye at sunset?

Re: What is the Linear speed of Sun in FE?
« Reply #3 on: September 29, 2017, 06:17:37 AM »
So at 52m/s speed will it create sonic boom??

Offline mtnman

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Re: What is the Linear speed of Sun in FE?
« Reply #4 on: September 29, 2017, 04:01:53 PM »
So at 52m/s speed will it create sonic boom??
A sonic boom, in space... No.

Also, a sonic boom is not created by something going continuously faster than the speed of sound. It's made when something breaks through the speed of sound (in an atmosphere).
« Last Edit: October 01, 2017, 07:34:12 PM by mtnman »