# Given two normals on the circumference of a circle, find the change in height between the two.

I am provided with two unit normals which lie on the circumference of a circle. I am not provided with the points on the circumference that these normals stem from. I do know that the distance between these two points along the $$x$$ axis is 1. I am supposed to use the fact that a unique unit normal corresponds to a unique point on the surface of a circle (the normal points directly out from the center). I have been able to make some progress using a diagram of the problem: Where $$q$$ is the value I want to calculate. From this my workings are:

$$\theta = \arccos(\frac{n_1 \cdot n_2}{\|n_1\|\|n_2\|}),\quad \gamma = \frac{(\pi - \theta)}{2},$$ $$\delta = \arctan(\frac{n_1.y}{n_1.x}),\quad \alpha = \gamma - \delta,\quad \beta = \frac{\pi}{2} - \alpha,$$ $$q = \frac{\sin(\alpha)}{\sin(\beta)}$$

A couple of simplifications can be made: $$\arccos(\frac{n_1 \cdot n_2}{\|n_1\|\|n_2\|})$$ is simply $$\arccos(n_1 \cdot n_2)$$, as both normals have magnitude 1. We can also say that $$q = \frac{\sin(\alpha)}{\sin(\beta)}$$ is really $$q = \tan(\alpha)$$, as $$\sin(90-\alpha) = \cos(\alpha)$$.

I'm only getting the correct results for the first quadrant of the circle. I'm not sure how to correct the method to give the correct value of $$q$$ for any two normals. I believe the problem is that I sometimes require the obtuse angle between the normals, but I am always getting the acute angle? For reference I am programming this problem, maybe I can use the atan2 function?

A second query is, do I actually need to use any trigonometric functions here? I think I can avoid the call to arctan when calculating $$\delta$$:

$$\tan(\alpha) = \tan(\gamma - \delta)\\ = \frac{\tan(\gamma)-\tan(\delta)}{1-\tan(\gamma)\tan(\delta)}$$ Where $$\delta=\arctan(\frac{n_1.y}{n_1.x})$$ so can I just avoid the $$\arctan$$?

Any tips would be great, thank you.

## 1 Answer

Let $$e^1=(1,0)$$ and $$e^2=(0,1)$$ be the standard orts and $$n=n_1-n_2$$. Then $$1=r |n\cdot e^1|$$ and $$q=r n\cdot e^2$$, so $$q=\frac {n\cdot e^2}{|n\cdot e^1|}$$.

• Hi, thanks for your answer, this seems like what I was looking for! before I accept this, could you add a little more explanation? Specifically, what is an "ort", I haven't come across that term before. Also how did you arrive at $1=r |n\cdot e^1|$ and $q=r n\cdot e^2$? Are they standard circle formulae? (sorry my understanding is quite basic). – nitronoid Jun 2 at 14:40
• @nitronoid By a standard ort $e^i$ in $\Bbb R^n$ I mean a vector whose $i$-th coordinate is $1$ and other coordinates are $0$. Indeed, this term may be not very common in English mathematical literature. The equalities follows from the next observation. Let $O$ be the center of the circle $A_1=(x,y)$, and $A_2=(x+1,y+q)$. Then $q$ is the projection of the vector $A_2A_1=OA_1-OA_2=rn_1-rn_2$ onto the $y$-axis (which equals $r(n_1-n_2)\cdot e^2$) and $1$ is the length of projection of the same vector onto the $x$-axis, that is $1=|r(n_1-n_2)\cdot e^1|$. – Alex Ravsky Jun 2 at 17:02
• Great! That makes perfect sense, thank you! – nitronoid Jun 2 at 19:27