# Math Help - Tangential Arc Problem

1. ## Tangential Arc Problem

Greetings all. I appreciate in advance any assistance you can provide. This is a real world problem for which I am able to construct a solution in a CAD system, however, I need to figure out how to determine this mathmatically. I am normally pretty confident with trig, but this one has me stumped. This has become a fairly common scenario in my line of work and neither myself, nor my coworkers have been able to figure out a solution without the use of snap tangent in a CAD system.

Reference the attached CAD drawing. Known values are B, D, W, X, Y, C1 and R1. Points A and C are arbitrary extensions of a line originating at B and D respectively. With the given information, I need to construct an arc which is tangent to arc XY and line CD.

I have been trying to determine a method that would give me the distance from D to Z, R2 and degrees of turn. However, I have not come up with anything even close to a workable solution. Again, any assistance is very much appreciated. Thanks.

2. Hello bcdave

Welcome to Math Help Forum!
Originally Posted by bcdave
Greetings all. I appreciate in advance any assistance you can provide. This is a real world problem for which I am able to construct a solution in a CAD system, however, I need to figure out how to determine this mathmatically. I am normally pretty confident with trig, but this one has me stumped. This has become a fairly common scenario in my line of work and neither myself, nor my coworkers have been able to figure out a solution without the use of snap tangent in a CAD system.

Reference the attached CAD drawing. Known values are B, D, W, X, Y, C1 and R1. Points A and C are arbitrary extensions of a line originating at B and D respectively. With the given information, I need to construct an arc which is tangent to arc XY and line CD.

I have been trying to determine a method that would give me the distance from D to Z, R2 and degrees of turn. However, I have not come up with anything even close to a workable solution. Again, any assistance is very much appreciated. Thanks.
Perhaps I'm not understanding the complexity of the problem here, but the solution seems to be extremely simple.

As I understand it, unless your diagram is a special case, XY is a diameter of the circle centre C1, radius R1, which is perpendicular to the line DC. You then need to find the centre, C2, and radius, R2, of the circle an arc of which is tangential to the semi-circle XY at its mid-point, and which is also tangential to the line CD.

Suppose that the distance of C1 from the line CD is x. Then, if my assumptions are correct, R2 = x and C2 lies on the line through C1 parallel to CD, where the distance C1C2 = R2 - R1 = x - R1.

What am I missing here?

Thank you very much for the quick response with an amazingly simple solution. You understood my problem perfectly. I was definitely over-thinking this one. Sometimes it just takes a set of fresh eyes to see the light.

While this works for my current scenario, where lines WX, AB and CD are all parallel, we do get a several cases where line CD is not parallel to line AB. Would you have any suggestions for when these line are not parallel? Thanks again.

(very glad I found/joined this forum, I will be visiting often)

4. Hello bcdave
Originally Posted by bcdave
...While this works for my current scenario, where lines WX, AB and CD are all parallel, we do get a several cases where line CD is not parallel to line AB. Would you have any suggestions for when these line are not parallel?...
Suppose that the line $DC$ makes an angle $\theta$ with $BA$, and that the perpendicular distance from $C_1$ to $DC$ is $d$. (I am assuming that this angle and this distance are known.)

Then it's fairly straightforward to show that the radius of the new arc, $R_2$, is given by:
$R_2= \frac{d-R_1\sin\theta}{1-\sin\theta}$
( Note that, where $DC$ and $BA$ are parallel, $\theta = 0$, and $R_2 = d$, as we found before.)

So the position of $C_2$ is along the line joining the 'tangent point' on your diagram (the mid-point of the semi-circle $XY$) to $C_1$, at a distance $R_2-R_1$ from $C_1$, where:
$R_2-R_1 = \frac{d-R_1}{1-\sin\theta}$
I haven't time now to draw the diagram and show you the working, but if you want to try it, find the point of intersection, $P$ say, of this 'line of centres' with $DC$. Then:
$PC_1 = \frac{d}{\sin\theta}$

$\Rightarrow PC_2 = \frac{d}{\sin\theta}+R_2-R_1$

$\Rightarrow R_2 = PC_2\sin\theta$
$=\left(\frac{d}{\sin\theta}+R_2-R_1\right)\sin\theta$
Solving for $R_2$ gives the result above.

Again, I am amazed and very grateful for your quick response. I will work on your solution this evening and respond back tomorrow. Thank you very much.

After working through your solution yesterday evening, I think I may have come up with simple solution for when line CD is not parallel to AB. Thought I'd run it by you.

Suppose the perpendicular distance from C1 to CD is d and this distance is also used as the R2 value. θ is the angular difference between AB and CD. Then the location of C2 could be found with d * tan(θ). This would place point Z where the line from C1 intersects CD, but I believe that would be acceptable and I would still get my required tangents to arc XY and line CD.

Does this sound correct to you? Thanks again.

7. Hello bcdave
Originally Posted by bcdave

After working through your solution yesterday evening, I think I may have come up with simple solution for when line CD is not parallel to AB. Thought I'd run it by you.

Suppose the perpendicular distance from C1 to CD is d and this distance is also used as the R2 value. θ is the angular difference between AB and CD. Then the location of C2 could be found with d * tan(θ). This would place point Z where the line from C1 intersects CD, but I believe that would be acceptable and I would still get my required tangents to arc XY and line CD.

Does this sound correct to you? Thanks again.
I'm not clear about what you mean here. Could you draw me a diagram?

I attach a diagram explaining my previous post. The essential fact I'm using is that the radius of a circle drawn through the point of contact of a tangent is perpendicular to the tangent. So:
$\angle PQC_1 = \angle PZC_2 = 90^o$
and, of course, the radius through S, the point of contact of the common tangent to the two circles, passes through $C_1$ and $C_2$.

The remainder of the working was in my previous post.

That makes perfect sense, especially after seeing your drawing. I have attached an additional two drawings. The first [Tangential Arc Problem 3] references my previous post.

If you suppose the perpendicular distance from C1 to line AC is d and θ is the angle at point A, then:
θ also equals the angle at point Z.
Therefore, the distance from C1 to C2 is given by d * tan(θ).
Simple Pythagorean theorem can now be used for the R2 value.
This would place point Z where the line from C1 intersects AC, but I believe that would be acceptable and I would still get my required tangents to arc XY and line CD. I was just wanting to verify that this sounds correct to you also.

The second drawing [Tangential Arc Problem 4] illustrates the last (and probably the most difficult) case I deal with. In this case, point Z is a known location and an arc must be constructed from a tangent point on arc XY to point Z tangent to line AC. The drawing represents the CAD solution drawn using snap tangents to show point C2 and R2. I have no idea how to even begin on a mathematical solution for this. I understand if this is too complex to get into on these forums and I don't want to be a bother, so if you are unable to assist with this last case, do you have someplace you could recommend I can go to online to review constructions such as this? Thanks again. You've been a great help.

9. Sorry, the second drawing was a little off. Please reference this drawing for "Tangential Arc Problem 4."

10. Hello bcdave
Originally Posted by bcdave

That makes perfect sense, especially after seeing your drawing. I have attached an additional two drawings. The first [Tangential Arc Problem 3] references my previous post.

If you suppose the perpendicular distance from C1 to line AC is d and θ is the angle at point A, then:
θ also equals the angle at point Z.
Therefore, the distance from C1 to C2 is given by d * tan(θ).
Simple Pythagorean theorem can now be used for the R2 value.
This would place point Z where the line from C1 intersects AC, but I believe that would be acceptable and I would still get my required tangents to arc XY and line CD. I was just wanting to verify that this sounds correct to you also.
I think there are two problems with this argument:

1. In your diagram (problem 3), $d$ is not the perpendicular distance from $C_1$ to $AC$; if it were, $\angle AZC_1$ would be $90^o$.
.
2. There is no reason that I can see why $\angle ZC_1C_2$ should be $90^o$ either. You have assumed this in giving the distance $C_1C_2$ as $d\tan\theta$.

The second drawing [Tangential Arc Problem 4] illustrates the last (and probably the most difficult) case I deal with. In this case, point Z is a known location and an arc must be constructed from a tangent point on arc XY to point Z tangent to line AC. The drawing represents the CAD solution drawn using snap tangents to show point C2 and R2. I have no idea how to even begin on a mathematical solution for this. I understand if this is too complex to get into on these forums and I don't want to be a bother, so if you are unable to assist with this last case, do you have someplace you could recommend I can go to online to review constructions such as this? Thanks again. You've been a great help.