# Internal angles of a triangle

Page 3 of 4
Goto page Previous  1, 2, 3, 4  Next

593352.  Fri Jul 31, 2009 5:45 pm

Posital wrote:
 Davini994 wrote: What was mentioned about the equivalence of a triangle with internal angles 5π and π on a globe is incorrect

If you have a very small triangle on a globe, the sum will tend towards π. But who is to say that the rest of the globe can't be considered a triangle with a approx 5π sum?

Well, if you were paying attention you'd know the answer to the last sentence is: nobody. This is where we got the 5π from.

And as you've said, a small triangle on an sphere will tend towards π; if it *is* π, then you must have a flat geometry (assuming consistent geometry), contrary to what you said earlier.

593358.  Fri Jul 31, 2009 6:08 pm

 Davini994 wrote: Well, if you were paying attention
Not any more. Contradict yourself all you want.

593380.  Fri Jul 31, 2009 7:20 pm

 ColinM wrote: The problem with spherical geometry is that you need a slightly different idea of what is meant by "line" and "point". A "point" is a pair of points on opposite sides of the sphere (antipodal points), and a line is a great circle, which has to go through the antipode of all its points (lines of longitude are great circles, but lines of latitude other than the equator are not).

Good point. Just to clarify, great circles in spherical geometry are equivalent to straight lines in Euclidean geometry. Nonequatorial lines of latitude can be considered as non-straight lines, i.e. they correspond to “curves” in Euclidean geometry.

Last edited by Susannah Dingley on Fri Jul 31, 2009 7:34 pm; edited 2 times in total

 593381.  Fri Jul 31, 2009 7:22 pm It's fairly easy to imagine a triangle on a sphere with internal angles of almost π: one with almost 2π and two with almost 3π/2.

637206.  Sun Nov 15, 2009 8:31 am

 Posital wrote: Ok - how about this. Sit at the pole - and wave to someone on the equator (say Libreville, Gabon). Now turn 90 degrees, and you'll see someone else on the equator (say Chone, Ecuador). The lines between the three of you make a triangle. But the angle between each of the lines is 90 degrees. Add them all up, and this triangle has 3x90=270 degrees internally. Which is more than the 180 degrees we might expect. That's why I didn't become a teacher. see: cantexplaingeometry.gov.uk

Sunday afternoon boredom has lead me to dig up old threads for entertainment and this post got me thinking.

While it is an excellent way of describing to the uninitiated how to imagine a triangle in spherical geometry with internal angles adding to 270 degrees (much as it pains me to abandon radians for the purpose), I find it QI that in order for you and your friends in Gabon and Ecuador to all be visible to one another, you would each need to be over 2500km above sea level.

I find this interesting only because it forces you out of the unconscious delusion of living on a (near enough for someone without a jet plane) infinite Euclidean plane. I live on a hill in London and on a clear day can see perhaps 15-20 miles away before local geographical features and curvature of the Earth deny me any more.

To ramble on a little further, I love to imagine what it would be like if Earth's gravity were far stronger and I were able to withstand it for long enough to see Australia (and indeed most of the Earth's surface) from the top of a single tall building.

637226.  Sun Nov 15, 2009 9:19 am

 spamimperial wrote: To ramble on a little further, I love to imagine what it would be like if Earth's gravity were far stronger and I were able to withstand it for long enough to see Australia (and indeed most of the Earth's surface) from the top of a single tall building.

Do you mean so strong that light wraps around from Australia?

 637231.  Sun Nov 15, 2009 9:26 am Exactly. The kind of gravitational field you'd get on the surface of a neutron star or similar. It would be a fascinating visual experience, or at least something I find impossible to imagine.

 637285.  Sun Nov 15, 2009 11:14 am Off the top of my head, I think that would only be possible inside an event horizon. I think it may be a perfect orbit *at* the event horizon. This is based on a Gollantz SF book, filtered through my memory; for a more reliable comment we may need Mr cscott.

637320.  Sun Nov 15, 2009 1:01 pm

Surely what I described would more likely apply to any surface about a mass from which the escape velocity, v obeys

0 << v < c .

Light is bent by the extreme gravitational field, but is still able to escape as long as it's aimed at a steep enough angle. The remainder of photons, unless they interact with something in the mean time, are left only with the option to bend away from a radial path and to eventually come back to ground far away. I'm sure the same thing would apply within or even at the event horizon, but it surely isn't necessary to be able to see Australia (not from the side at least, from above would be tricky).

PS - A quick glance at Wikipedia would appear to agree with me (http://en.wikipedia.org/wiki/Neutron_star#Properties).

 Quote: The gravitational field at the star's surface is about 2 × 10^11 times stronger than on Earth. The escape velocity is about 100,000 km/s, which is about one third the speed of light. Such a strong gravitational field acts as a gravitational lens and bends the radiation emitted by the star such that parts of the normally invisible rear surface become visible.

Edit - Intuition has got the better of me in assuming that direction was at all related to escape velocity for photons. While the Earth's atmosphere might cause an object to escape when pointed vertically but not otherwise, it would certainly have no such effect on light. I am now forced to question my entire thought process above, and indeed mine and Wikipedia's claims about neutron stars.

637351.  Sun Nov 15, 2009 3:25 pm

 Spamimperial wrote: Intuition has got the better of me in assuming that direction was at all related to escape velocity for photons.

Very, very interesting.

 Quote: 0 << v < c

A counterexample for this is the earth, as light travelling horizontally escapes of course.

What wiki is talking about is light from the object bending slightly on its path to Earth, so a little bit more is visible. It's not the back though, that's bending of a very different scale. It's because it's bending round the object you are standing on, so it has to pretty much be an orbit.

This has more discussion on escape speed, as we should call it now!

http://astroprofspage.com/archives/1554

 637354.  Sun Nov 15, 2009 3:35 pm I feel silly for making such a mistake and then overthinking the result. This is the first November since 1990 that I haven't been in full-time education and it's obviously taking it's toll already. I also feel rather disappointed that even in my own thought experiment, I can't see Australia without having to go there myself.

 637362.  Sun Nov 15, 2009 3:46 pm You could always go to Google Earth.

638941.  Sat Nov 21, 2009 7:56 am

 ColinM wrote: Lets have a go at explaining some of it then............ Now consider a triangle on the Earth. It has one point at the north pole, and the other two on the equator. Two of its lines are lines of latitude and the other is the equator, so they're all legitimate lines. What are the angles at each corner? Well, every line of latitude meets the equator at a right angle, or 90 degrees. We have two of those, so that's 180 degrees, or π radians. But our last angle can be whatever we like - we can pick anything from almost nothing to nearly a full circle, and still make a triangle. The sum of the angles will thus be more than π (or 180 degrees), contradicting our theorem in plane geometry. Does that help anyone at all?

Yes, that was helpful, except that I think you meant "longitude" where you typed "latitude"

Just to complete your nice explanation, here's a picture from Wiki.

 769744.  Sat Dec 25, 2010 5:04 pm To clarify a bit: In plane geometry, a triangle is a figure consisting of three line segments joined at their ends. The angles formed by the three intersections add up to 180 degrees. A spherical triangle is a figure on the surface of a sphere, consisting of three segments of great circles of that sphere joined at their ends. The angles formed by the three intersections add up to an angle between 180 degrees (representing an infinitesimal triangle) and 540 degrees (representing a great circle that coincides with the other three). A great circle, by the way, is a circle on the surface of a sphere whose center coincides with the center of the sphere -- called a great circle because it is the largest circle that can be drawn on that sphere. rj

769833.  Sun Dec 26, 2010 7:59 am

 Spamperial wrote: I also feel rather disappointed that even in my own thought experiment, I can't see Australia without having to go there myself.
Don't lose heart - with a number of extra-planetary objects you could slowly bend light back to earth without needing any event horizons!

But you'd have to accept a very very poor image quality.

You could also opt for "seeing" it in radio frequency - and hope for very unusual ionospheric conditions. Perhaps listen to Radio Brisbane... and join the Summer FUNomenon at Dreamworld...

The QI options are endless...

Page 3 of 4
Goto page Previous  1, 2, 3, 4  Next

All times are GMT - 5 Hours

Display posts from previous:

Forum tools
User tools