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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.

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In summary, the metric for a 2-sphere can be expressed as $$ds^2 = dr^2 + R^2sin(r/R)d\theta^2$$ and the area of a triangle on the sphere can be calculated by using spherical trigonometry or by using the more familiar metric, $$ds^2 = R^2 ( d^2 \theta + \sin^2(\theta ) d^2 \phi )$$ with ranges of ##0\leq r \leq \pi R/2## for the first term and ##0\leq \theta \leq \pi## and ##0\leq \phi \leq 2\pi## for the second term.

- #1

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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.

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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

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I did not understand what you are trying to mean.jeff davis said:

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

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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.

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

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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?Arman777 said: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.jeff davis said:I think that the area of a triangle in metric would be the same as in imperial.

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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 asOrodruin said:It is unclear what you mean. You obviously need some information about the triangle itself

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

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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)##Orodruin said:

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

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

$$ds^2 = dr^2 + R^2sin(r/R)^2d\theta^2$$ 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

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

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

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Again, this is not a triangle. It is a circle sector.Arman777 said:I ma kind of having trouble to understand this. At post 11 $$h = ARsin(r/R)$$ ? I mean it has to be that way..

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

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Of the circle sector or the triangle?

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TriangleOrodruin said:Of the circle sector or the triangle?

- #19

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Arman777 said: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?

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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...

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