- #1

Arman777

Gold Member

- 2,165

- 185

Is there an equation to describe the area of an triangle by using metric.

Note: I know the formulation by using the angles but I am asking for an equation by using only the metric.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter Arman777
- Start date

- #1

Arman777

Gold Member

- 2,165

- 185

Is there an equation to describe the area of an triangle by using metric.

Note: I know the formulation by using the angles but I am asking for an equation by using only the metric.

- #2

jeff davis

- 55

- 13

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.

jeff

- #3

Arman777

Gold Member

- 2,165

- 185

I did not understand what you are trying to mean.

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.

jeff

- #4

jeff davis

- 55

- 13

are you referring to this? I have never tried to solve for the spherical area of a triangle before. It is interesting.

I did not understand what i meant either. LOL.

- #5

Arman777

Gold Member

- 2,165

- 185

Well kind of. But I understand the problem ...so I guess there's no need

- #6

- 20,004

- 10,647

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.

- #7

Arman777

Gold Member

- 2,165

- 185

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 that's possible.. Let's 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

$$Area = \int_0^A \int_0^r Rsin(r/R)drd\theta$$

Last edited:

- #8

- 20,004

- 10,647

- #9

Arman777

Gold Member

- 2,165

- 185

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)##

Actually the area is $$Area = \int_0^rARsin(r/R)dr$$ but this is also equal to $$Area = \int_0^A \int_0^r Rsin(r/R)drd\theta$$

- #10

- 20,004

- 10,647

- #11

Arman777

Gold Member

- 2,165

- 185

Hmm I know that ##r ∈[0, \pi R/2)##. But that's confusing. I mean

$$ds^2 = dr^2 + R^2sin(r/R)^2d\theta^2$$ so for ##h##, ##dr = 0 (?)##.

I see you are meaning that if ##r\ne \pi R/2## then ##dr\ne0##

If we know the value of the A is it possible to find the h from sherical trigonometry (without knowing the value of ##r##)? I guess metric is useless to find the ##h## then

- #12

- 20,004

- 10,647

- #13

Arman777

Gold Member

- 2,165

- 185

Hmm I see. I ll try to look something else for solution. Thanks then

- #14

Arman777

Gold Member

- 2,165

- 185

I ma kind of having trouble to understand this. At post 11 $$h = ARsin(r/R)$$ ? I mean it has to be that way..line with the bottom length you have specified is not a geodesic unless rrr is the distance from the pole to the equator.

- #15

- 20,004

- 10,647

Again, this is not a triangle. It is a circle sector.I ma kind of having trouble to understand this. At post 11 $$h = ARsin(r/R)$$ ? I mean it has to be that way..

- #16

Arman777

Gold Member

- 2,165

- 185

Okay.. umm then how can I find the area of that thing. What should I do ? Is it better maybe to use spherical coordinates ?Again, this is not a triangle. It is a circle sector.

- #17

- 20,004

- 10,647

Of the circle sector or the triangle?

- #18

Arman777

Gold Member

- 2,165

- 185

TriangleOf the circle sector or the triangle?

- #19

DEvens

Education Advisor

Gold Member

- 1,203

- 460

The metric for 2-sphere is $$ds^2 = dr^2 + R^2sin(r/R)d\theta^2$$

Here, by ##r## do you mean ##R \theta##? Then your metric would be the more familiar form.

$$ds^2 = R^2 ( d^2 \theta + \sin^2(\theta ) d^2 \phi )$$

And ##\theta## would have range 0 to ## \pi## and ##\phi## would have range 0 to ##2 \pi##. Then it all becomes very familiar surface-of-a-sphere type geometry. Is that it?

- #20

Arman777

Gold Member

- 2,165

- 185

But I used the spherical trigonometry to find the area of the triangle.. But the metric you wrote also seems well for my problem as well. Sadly I already handed out the paper...

Share:

- Last Post

- Replies
- 8

- Views
- 484

- Replies
- 10

- Views
- 3K

- Last Post

- Replies
- 16

- Views
- 1K

- Last Post

- Replies
- 13

- Views
- 445

- Last Post

- Replies
- 13

- Views
- 2K

- Last Post

- Replies
- 6

- Views
- 963

- Last Post

- Replies
- 3

- Views
- 3K

- Replies
- 4

- Views
- 505

- Last Post

- Replies
- 5

- Views
- 718

- Last Post

- Replies
- 1

- Views
- 1K