Rotation formula

This question came up at a school star party last night.

I'm sure there's a fairly straightforward formula for this (probably
something involving sines, cosines and other things I've forgotten that I
ever knew)... but I can't seem to find it. Maybe I just don't know the right
question to ask.

Assuming (in rough round numbers) that the Earth's diameter at the equator
is 24,000 miles, then someone standing at the equator is moving 24,000 miles
per hour, right?

How do I determine the speed for another latitude. We happen to be at 21.5
deg. N. I know that if I was at 45 deg. N. the diameter of the Earth would
not be 1/2 of 24,000 miles... but just how much would it be?

Kilolani wrote:
Assuming (in rough round numbers) that the Earth's diameter at the equator
is 24,000 miles, then someone standing at the equator is moving 24,000 miles
per hour, right?

Not quite. The Earth's *circumference* (not diameter) is just about
25,000 miles, and it rotates once in about 24 hours. Thus, the linear
speed at the equator is a bit over 1,000 miles per hour.

How do I determine the speed for another latitude. We happen to be at 21.5
deg. N. I know that if I was at 45 deg. N. the diameter of the Earth would
not be 1/2 of 24,000 miles... but just how much would it be?

Thank goodness. I thought you were going to ask the formula for field
rotation. The linear speed at any latitude L can be given as

V = Ve * cos L

where V is the speed you want, and Ve is the speed at the equator.

Brian Tung
The Astronomy Corner at http://astro.isi.edu/
Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/
The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/
My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.txt

On Sun, 14 Mar 2004 06:01:25 GMT, "Kilolani"
wrote:

This question came up at a school star party last night.

I'm sure there's a fairly straightforward formula for this (probably
something involving sines, cosines and other things I've forgotten that I
ever knew)... but I can't seem to find it. Maybe I just don't know the right
question to ask.

Assuming (in rough round numbers) that the Earth's diameter at the equator
is 24,000 miles, then someone standing at the equator is moving 24,000 miles
per hour, right?

You are traveling 24,000 miles per day, or 1000 miles per hour.

How do I determine the speed for another latitude. We happen to be at 21.5
deg. N. I know that if I was at 45 deg. N. the diameter of the Earth would
not be 1/2 of 24,000 miles... but just how much would it be?

It is just the speed at the equator times the cosine of the latitude. So at 21.5
degrees, that is (1000 mph)*(0.93)=930 mph.

_________________________________________________

Chris L Peterson
Cloudbait Observatory
http://www.cloudbait.com

This question came up at a school star party last night.
How do I determine the speed for another latitude. We happen to be
at 21.5
deg. N. I know that if I was at 45 deg. N. the diameter of the Earth would
not be 1/2 of 24,000 miles... but just how much would it be?



I will take a crack at this one. interesting problem.


v=(omega)*r. where omega is angular
velocity in radians. Since your angular velocity (omega) is the same
no matter where you are, the problem simplifies quite a bit.

period=(2*pi)/omega, ... (2*pi/24hours) = omega


so,

v= (2*pi/24hours)* (radius)



The radius isn't difficult to find.

(The diameter of earth is 7,926 miles)

so the radius is probably something like, 3,963*cos(degrees lat)


so,


v= (2*pi/24hours)* 3963 *cos(21.5)


Speed at the equator is 1037.5 mph

and at, 21.5 degrees. ... 965.3 mph

I hope this is right.

"Brian Tung" wrote in message
...
Kilolani wrote:
Assuming (in rough round numbers) that the Earth's diameter at the
equator
is 24,000 miles, then someone standing at the equator is moving 24,000
miles
per hour, right?

Not quite. The Earth's *circumference* (not diameter) is just about
25,000 miles, and it rotates once in about 24 hours. Thus, the linear
speed at the equator is a bit over 1,000 miles per hour.

oops... yes, sorry. I meant circumference, and I meant 24,000 miles per day
(1,000 mph)

How do I determine the speed for another latitude. We happen to be at
21.5
deg. N. I know that if I was at 45 deg. N. the diameter of the Earth
would
not be 1/2 of 24,000 miles... but just how much would it be?

Thank goodness. I thought you were going to ask the formula for field
rotation. The linear speed at any latitude L can be given as

V = Ve * cos L

where V is the speed you want, and Ve is the speed at the equator.

Thanks, Brian... I knew you would know the answer!

"Chris L Peterson" wrote in message
...

It is just the speed at the equator times the cosine of the latitude. So
at 21.5
degrees, that is (1000 mph)*(0.93)=930 mph.

Thanks Chris... I now know that I shouldn't have scoffed in math class when
they told me trigonometry would be useful someday!

I'm sure there's a fairly straightforward formula for this

Hi:

Yup:

V = 1671(cosL)

V = rotational velocity in kph

L = latitude

Just plug the above into a scientific calculator (including the one in
Windows), and you'll have your answer.


Peace,
Rod Mollise
Author of _Choosing and Using a Schmidt Cassegrain Telescope_
Like SCTs and MCTs?
Check-out sct-user, the mailing list for CAT fanciers!
Goto http://members.aol.com/RMOLLISE/index.html

"Kilolani" wrote in message thlink.net...
This question came up at a school star party last night.

I'm sure there's a fairly straightforward formula for this (probably
something involving sines, cosines and other things I've forgotten that I
ever knew)... but I can't seem to find it. Maybe I just don't know the right
question to ask...

Yes. Think about the problem:

You're looking for a function to do with circles
whose value is 1 at zero degrees, and 0 at 90 degrees.
Sounds like a cosine. If it's going to be anything
more complicated, you'll need to explain why.

You could also draw a diagram.

Laura Halliday VE7LDH "Que les nuages soient notre
Grid: CN89mg pied a terre..."
ICBM: 49 16.05 N 122 56.92 W - Hospital/Shafte

"Kilolani" wrote in message thlink.net...
Assuming (in rough round numbers) that the Earth's diameter at the equator
is 24,000 miles,

It is 24,000 Nautical Miles not statute miles

Mitch Alsup wrote:
It is 24,000 Nautical Miles not statute miles

Originally, the nautical mile was defined as 1/60 of a degree at the
Earth's equator. That would mean that the Earth's circumference would
be 360*60 or 21,600 nautical miles, by definition.

Because of various lumpinesses (how's that for an awkward word?) in
the shape of the Earth, the nautical mile was redefined in 1929 to be
exactly 1,852 meters. Among other things, this means that the speed
of light is *exactly*

161874 and 905/926 nautical miles per second

File under useless!

Brian Tung
The Astronomy Corner at http://astro.isi.edu/
Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/
The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/
My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.txt