Help on the freaking simple question that is ****ing me off at the moment

"Ron Miller" wrote in message
...

"Zarkovic" wrote in message
news:ye9ab.8990$CU3.7779@pd7tw3no...
Hi guys, here is one question that I can't get for some reason, I just
can't. I come close to it, but everytime I am by 300km/hr off. Anyhow
here
is the question and could someone please tell me how it's done.

"Derive the equation from which you could calculate speeds of someone on
the
surface of the Earth, depending on their latitude. The equation will
depend
on person's latitude, (Q), the radius of the Earth, (Re), and the period
of
rotation, (T). ( I tried it, but nothing.) (Hints: the speed of a person
on
a Northpole is 0 km/hr and speed of a person on the Equator is 1650
km/hr)


I don't know what the radius of the earth would have to do with it. Your
speed depends on your perpendicular distance from the earth's axis (as
though the latitude you are at is the circumference of a wheel that is
spinning on the axis).
Because that perpendicular distance, is cos(Q)*Re...
You either have to have the radius of the Earth, or another way of working
out that perpendicular distance.

Best Wishes

I'm thinking it might be 1650*cos(lat)

I'm thinking it might be 1650*cos(lat)

"Zarkovic" wrote in message
news:XR9ab.9029$CU3.1539@pd7tw3no...
Ahhh man, thanks guys. For some reason I turned my sphere around and I
used
for equator speed cos (90) which equals to 0 , but I should have used cos
(0) instead. Sorry for the mix up, and thanks once again to both of you.



It is nice to have people around when when the answer and the solution are
not in the back of the textbook. )

todd

"Zarkovic" wrote in message
news:XR9ab.9029$CU3.1539@pd7tw3no...
Ahhh man, thanks guys. For some reason I turned my sphere around and I
used
for equator speed cos (90) which equals to 0 , but I should have used cos
(0) instead. Sorry for the mix up, and thanks once again to both of you.



It is nice to have people around when when the answer and the solution are
not in the back of the textbook. )

todd