How do you figure the circumference of a sphere at different points?

Question

Assume a sphere where you know the diameter at the 'equator'. How do you go about finding the diameter at the "tropic of cancer" (the half-way point between the equator and the north pole)? Logic would say that it is 50% of the biggest diameter, but when I draw it out, that doesn't seem to be the case.

Answer 1

if you know the angle from horizontal

the circumference is the cosine of this angle

for 50% cos45=x / R

x = R cos 45 = .707 R

circum = 2 pi x

alternate method

a^2 + b^2  = c^2

a=x horz. dist from center

b=y vertical dist from center

c = radius

x^2 + y^2 = r^2

at 45deg x=y then

x^2 +x^2 = r^2

2x^2 = r^2

x= sqrt2/2 r = 0.707 r

Answer 2

actually logic would not show 50%... what you do is draw your tropic of cancer, then where the line hits your circle, you draw two radius's of the circle... and since you know the diameter, the radius will of course be half of that, then you use pythagorius theorom,   a^2 + b^2 = c^2.. and this will tell you the length of it.

Answer 3

Well if you draw a cut away of the sphere, the radius from the center of the sphere to a point on the tropic of cancer is 1 (as well as any other radius).  If you drop an altitude to your radius for your diameter (so that the segment joins the point on the tropic of cancer and your diameter at a right angle), you make a triangle with angle measures of 45-45-90

this means that each side of the triangle is 1/(√2) if you revolve this length around your diameter, then you would trace out the tropic of cancer, so use the formula for circumference to find the circumference

c=2πr

c=2π(1/(√2))

c=2π/(√2)

if you want a proper fraction, c=π√2

comparatively: the circumference is 2πr at the diameter, or 2π with a diameter of 1.  This is about 6.28.  The diameter of the tropic of cancer is π√2 which is about 4.44.  As a percentage: 4.44/6.28 is about  71% so the tropic of cancer is not 50%, as you saw when you drew it

You can use trig and drawing a triangle to find the circumference at any latitude  

Answer 4

For all points on a circle whose radius is R, X = SQRT(R^2 - Y^2). If X is the new radius and Y0 is essentially the lattitude, then the circumference of the new circle would be SQRT(R^2 - Y0^2) and the circumference would be 2*PI*SQRT(R^2 - Y0^2). For the halfway point, Y0 = R/2 and substitution yields

X = SQRT(R^2 - (R/2)^2)

X = SQRT(R^2 - R^2/4)

X = SQRT(3/4R^2)

If the original radius was 10, then the radius of the halfway point would be SQRT(75) = 8.66 which is 86% of the original radius, not 50%

BTW: logic suggests the world is flat and the entire universe revolves around it.

Answer 5

radius=R*cosine(degrees north av equater)