I did not understand what you are trying to mean.Hello,
I think that the area of a triangle in metric would be the same as in imperial. It is just a ratio really. 1/2 B * H
would be the same for metric right?
unless you take the sector and neglect the segment.; Triangle = sector - segment. You could probably derive this in radians easily enough.
It is unclear what you mean. You obviously need some information about the triangle itself. Which information about the triangle do you want to use? Or are you after a more general integral expression for the area of a part of the sphere?Is there an equation to describe the area of an triangle by using metric.
This is not about units, it is about the metric tensor used in differential geometry as applied to a curved surface.I think that the area of a triangle in metric would be the same as in imperial.
Yes I realized that after some time later. First I thought that from the metric we can define a general area equation for any spherical triangle. However as you mentioned thats possible.. Lets suppose we have a triangle like this. N is the north pole and it makes an angle of A. From the metric I find that the angle can be expressed asIt is unclear what you mean. You obviously need some information about the triangle itself
Well yes sorry the other length is also r. So the bottom length becomes ##ARsin(r/R)dr## (from ##ds = Rsin(r/R)d\theta)##Your triangle is not well defined, you need to specify three numbers to completely specify the triangle.
Hmm I know that ##r ∈[0, \pi R/2)##. But thats confusing. I meanThat is not a triangle. A triangle has geodesics as its sides and the line with the bottom length you have specified is not a geodesic unless ##r## is the distance from the pole to the equator.
Here, by ##r## do you mean ##R \theta##? Then your metric would be the more familiar form.The metric for 2-sphere is $$ds^2 = dr^2 + R^2sin(r/R)d\theta^2$$