Is Our Weight Problem Worse than We Think?

I've been 10½ stone for the last twenty years.

😏

Here's an interesting "application" of this idea.

http://www.space.com/scienceastronomy/060110_vega_cool.html

Originally posted by Suzianne
It would have to be inversely proportional to the latitude.

As a function of the latitude, tho, I'm not at all sure about the actual angle of the tilt.

EDIT: (And yes, I did take trig in school, but I forget most of it, hehe)

I haven't done the calculations yet but I don't think it would be quite be inversely proportional. More like:

F = ((v^2)/r) cos (l)

where l is latitude.

Will try and verify this and work out the angle of tilt relative to gravity if the earth stops moving. 🙂

Ok, have confirmed that it:

Starting from F = (v^2)/r I decided to represent the equation in terms of F, r and t to make things a little easier.

d = 2*pi*r

therefore F = ((2*pi*r/t)^2)/r = (4 * pi^2 * r) / (t^2)

r would be proportional to cos(l) therefore

F = ((4 * pi ^ 2 * r) / (t^2)) cos (l)

Which effectively does confirm that F = ((v^2)/r) cos (l)

Agree that it is frustrating not being able to represent equations properly on here.

Still working on the tilt.

Diagram: http://www.lunacy.force9.co.uk/images/trig.jpg

My vector addition is a little rusty but I think I got it (and referring to my very badly done drawing). After resolving:

latitute = l = theta

a = angle of tilt = alpha

g = force of gravity

z and p are just variables I used while deriving.

h = F sin (l)

p = F cos (l)

z = F cos (l) + g

tan (a) = (F sin (l)) / (F cos (l) + g)

a = arctan ((F sin (l)) / (F cos (l) + g))

I haven't checked this through but I believe it is correct. I am also sure it can be simplified further but haven't bothered to go that far.

Edit: Forgot to substitute in F as in my previous post.

F = ((4 * pi ^ 2 * r) / (t^2)) cos (l)

a = arctan (((4 * pi ^ 2 * r) / (t^2)) cos (l) sin (l) / (((4 * pi ^ 2 * r) / (t^2)) cos^2 (l) + g))

I will simplify it tomorrow, getting late. 😛