# Thread: Arch Curve Drawing

1. ## Arch Curve Drawing

I want to confirm this, i want to draw an arch, knowing its shape, which is the parabolic equation ax^2+bx+c, knowing its end and final point and its lenght, not the coord, the real length of the arch (all these pararmeters it is me that decide them), what i think is if you know the starting and end points and you the length, knowing that is can only assume a specific shape you only have two solutions, one is the mirror of the other, however i cant define or the radius or the arch equation, i am hoping someone can enlighten me about this.
thank you

2. Originally Posted by RPinto2
I want to confirm this, i want to draw an arch, knowing its shape, which is the parabolic equation ax^2+bx+c, knowing its end and final point and its lenght, not the coord, the real length of the arch
The arclength of the parabola? That's going to be hard!

(all these pararmeters it is me that decide them), what i think is if you know the starting and end points and you the length, knowing that is can only assume a specific shape you only have two solutions, one is the mirror of the other, however i cant define or the radius or the arch equation, i am hoping someone can enlighten me about this.
thank you
Are the two endpoints you have at the same height so the arch is symmetric? That is, are the two endpoints (a, c) and (b, c)? You can simplify a little by letting d= (a-b)/2, half of the distance from a to b and then looking at a parabola from (-d, 0) to (d, 0). Then you know that the $y= ax^2+ b$ and the fact that y= 0 when x= d means $ad^2+ b= 0$ so that $b= -ad^2$.

Now the hard part. If $y= ax^2+ b$, then $y'= 2ax$ and the "differential of arclength" is $\sqrt{1+ 4a^2x^2}dx$. The arclength, from 0 to d, is $\int_0^d\sqrt{1+ 4a^2x^2}dx$. To do that, let $2ax= tan(\theta)$ so that $\sqrt{1+ 4a^2x^2}= \sqrt{1+ tan^2(\theta)}= \sqrt{sec^2(\theta)}= sec(\theta)$. Also $2a dx= sec^2(\theta)d\theta$. When x= 0, $\theta= 0$. When x= d, $\theta= arctan(2ad)$. The arclength is $\int_0^{arctan(2ad)}sec^3(\theta)d\theta= \int_0^{arctan(2ad)}\frac{d\theta}{cos^3(\theta)}$.

That, I think, can be done by a trig substitution followed by "partial fractions". In any case, set the arclength (from -d to d so twice that integral) equal to whatever the length is to be and solve for a and b.

3. thank you i will try, and yes they are at the same height, so the arch is symetrical