Question about proof that all surfaces with zero Gaussian curvature are developable
Hey guys, I need your help in understanding a proof given by Struijk on the claim that all surfaces with zero Gaussian curvature are developable surfaces. I am an engineering student rather than a mathematician, so excuse me if this is a complete beginners-question :)
Basically, the part of the proof I do understand is:
means that the determinant of the second fundamental tensor must be zero .
We can rewrite this
We obtain the following requirement for zero gaussian curvature: .
Now we have two possibilities:
2. is collinear with , which basically implies that case 1 will be true if we change to different coordinates
So far so good, but now the part that I don't understand. If we look at case 1 only, how does for example imply developability, meaning that the normal vector does not change along a straight line on the surface (generator)? Why must this be true and why can't the -curve be curved?
Thanks in advance!
Re: Question about proof that all surfaces with zero Gaussian curvature are developab
Make that .