# Formula for perimeter of projected figure

Given a surface $$S$$ in $$\mathbb{R}^3$$, we can calculate the area of its projection $$S'$$ onto a given plane $$P$$ using the formula $$[S'] = \iint_{S} \cos\beta \space dA$$, where $$\beta$$ is the angle between the surface and $$P$$ at a given point on $$S$$.

Is there a corresponding formula for the perimeter of a surface projected onto a plane? It seems like it would be messier, since not all lengths are scaled equally under projection.

• What about $\int_{\partial S}\cos\beta\,ds$? – Aretino Jul 18 at 17:07
• I don't think the above formula works. In the case that $S$ is a polygon that lies on a single plane $Q$, then if you have a side $AB$ of $S$ that lies parallel to the plane $P$, the projected length of $AB$ will equal its original length, even if $\beta \neq 0$. – yojan_sushi Jul 18 at 18:13
• Of course in this case $\beta$ should represent the angle between between the line and $P$ at a given point. – Aretino Jul 18 at 19:46