jonthebaptist
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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 [tex]\sigma\left(t,\varphi\right)=\left(\begin{array}{ccc}<br />
r\cos t, & r\sin t, & \omega t+\varphi\end{array}\right)[/tex]
I solve for the terms in First Fundamental Form, [tex]E=r^{2}+\omega^{2}[/tex]
[tex]F=\omega[/tex]
[tex]G=1[/tex]
Solving the surface area integral [tex]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.[/tex]
I solve for the terms in First Fundamental Form, [tex]E=r^{2}+\omega^{2}[/tex]
[tex]F=\omega[/tex]
[tex]G=1[/tex]
Solving the surface area integral [tex]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.[/tex]
