- #1
jonthebaptist
- 17
- 0
I am solving for the surface area of a helical ribbon that I represent as a ruled surface, the curve being the helix and the rulings being in the vertical \sigma\left(t,\varphi\right)=\left(\begin{array}{ccc}<br />
r\cos t, & r\sin t, & \omega t+\varphi\end{array}\right)
I solve for the terms in First Fundamental Form, E=r^{2}+\omega^{2}
F=\omega
G=1
Solving the surface area integral A=\iint\sqrt{EG-F^{2}}dtd\varphi[\tex]<br /> <br /> I get an area equal to the product of the total angular rotation times the radius times the width of the ribbon. This is the area of a rectangle with a length equal to the circumference of the circle that is projected by the helix onto the x-y plane. What perplexes me is that I would think that if this ribbon was isometric to a rectangle it would be one with a length equal to the arc-length of the helix, not the projected circle.
I solve for the terms in First Fundamental Form, E=r^{2}+\omega^{2}
F=\omega
G=1
Solving the surface area integral A=\iint\sqrt{EG-F^{2}}dtd\varphi[\tex]<br /> <br /> I get an area equal to the product of the total angular rotation times the radius times the width of the ribbon. This is the area of a rectangle with a length equal to the circumference of the circle that is projected by the helix onto the x-y plane. What perplexes me is that I would think that if this ribbon was isometric to a rectangle it would be one with a length equal to the arc-length of the helix, not the projected circle.
