I hope I can explain this correctly, and I do have a picture of what I’m trying to do.
From the diagram this is what I’m trying to figure out. The odd symbol after the R, C2, and I is supposed to be a ?, but doesn’t come up in Microstation.
There is a known circle, with known Radius and center point, R and C1. There is also a known point in space, P, and a known line L. What I’d like to do is find the circle that goes through the unknown point I? and through P. The R? line from P is perpendicular to L.
What I’ve been doing is making a circle off of L running it down R? until it just touches around I?. This works, but is hard to make exact and sometimes hard to see. If I could find one of the three unknowns then making the circle would be easy. Is there an easier way to do this?
Point P and line L are given, so that defines the dashed line in the diagram. We know C2 must be on that line. You just need to find a point on this line where the distance P-C2 is the sum of R and C1-C2. Does that help?
The answer also depends on if you want the answer in terms of coordinates, or a “compass and straight edge” solution.
A compass and straight edge approach is fine, the coordinates the computer can give. I think I half understand what you’re saying, but I’ll have to actually think it over to make sure.
It seems to me there are an infinite number of answers - it all depends on your R? (the second radius). L, the tangent, is somewhat irrelevant as it will always be perpendicular to your R?. So the only trick is finding C2? (the center of the other circle.
Given your initial circle (R and C1), pick a point for I. The distance from I to C2 will be the same as from P to C2 since they will have the same radius. Pick an R2 and you can then determine the center C2. (Note: C1 may be the same point as C2 depending on your R2).
Now if you want your initial circle (R, C1) to share only the point I, then, make sure C1 is on the line between I and C2.
But the thing is you can have any infinite number of R2s that will satisfy these criteria so long as R2 > R.
It’s been a while since I’ve done “compass and straight edge” geometry but I think you first draw a line (L2) perpendicular to L, and passing through P. Then define a point (P2) on that line that is distance R away from P. Now draw a line that bisects P2 and C1 (by defining 2 arbitrary points that are equidistance from P2 and C1). This line intersects L2 at C2.
In terms of coordinates, line L2 is
(y[sub]C2[/sub] - y[sub]P[/sub]) = a (x[sub]C2[/sub] - x[sub]P[/sub])
where a is the slope of line L. The other condition is
I understood the OP to mean P, L, C1 and R are all given.
L is not irrelevant. If the second circle is tangent to L at P, that means C2 must be on a line that is perpendicular to L and passes through P. You can still draw an infinite number of circles with the center on this line and is tangent to L, but only one of them (or none at all) will be also tangent to the original circle.
Yes, P, L, C1 and R are all known. What is actually happening here is that a plane will come around the first circle to I?, be passed off to the second circle to P and then fly a straight line L. It’s finding that small arc and pass off point that can’t be figured out without eyeballing it.
I’m using microstation as we speak so feel free to ask questions if this doesn’t do it for you.
I started with an arbitrary C1 circle and an arbitrary line L. I picked a random point on L and drew a line to it just so I would have a way to pick it this P point later.
I put the place circle tool in ‘edge’ mode, as opposed to center, and now of course you need to give it three points to define the circle.
I set my snap mode to tangent, clicked on the C1 circle and accepted it.
Set snap to key point and snap to P wherever that is - like I said, I drew a dummy line to L with the endpoint at P so I’d have something to snap to.
Set snap back to tangent and snap to the line L and there’s your circle. Or Bob’s your uncle, depending on your preference.
You might have to jack with the mouse a bit to get it drawing roughly the right circle (dynamically) before you can set the third tangent snap point.
When I tried going tangent, tangent, point P, I didn’t have any luck, but going tangent, P, tangent works just fine.
If I understand the question, which I may well not, there are an infinite number of curves that are tangent to both the known C1 circle and known line L - but only one of these is tangent to line L at known point P.
I’ll have to wait until Monday to try it out. We don’t do a whole lot of fancy stuff with Microstation, but I think I get the idea. The snap tangent was something I didn’t even know about.
You don’t have enough information to uniquely specify the second circle. If you were given both R? and I?, there would be at most one solution, but even having only one of them allows for infinitely many solutions. Are there any additional constraints?
As I see it, there are only two possible unknown circles. One just touching the known circle on the side near to P, the other encircling the known circle and touching it on the far side, as in the diagram.
Although the OP didn’t explicitly say so, from the diagram I think that the unknown circle must intersect the known one at exactly one point.
Call the line perpendicular to L, L’.
Let Q be a point on L’ that is distance t from P in the appropriate direction.
Let A be the measure of angle QPC1 and let d be distance PC1.
We’d like to find the value of t such that Q is the point we’re interested in. That occurs when distance QP is equal to distance QC1 plus R. (i.e. the minimum distance from Q to L equals the maximum distance from Q to the known circle.)
Using the law of cosines we get:
t = Sqrt(d^2 + t^2 -2dt cos(A)) + R
Solving for t yields:
t = (d^2 - R^2)/(2R + 2d cos (A))
Everything on the right hand side is known so we now have the radius of the larger circle. (Modulo a finite number of algebra mistakes that I may have made.)
Ok, two things there was indeed an algebra mistake in the above. It should be:
t = (d^2 - R^2)/(-2R + 2d cos (A))
I think.
But I now have a straightedge and compass style construction.
Define L’ as above. Let Q be a point on L’ such that distance QP equals R. Construct the perpendicular bisector of QC1. The intersection of this line and L’ is the point you want.
I think with my previous post that that image is self explanatory, but feel free to ask questions if it is not or I have incorrectly understood the problem.
This also means that the line QP is the same as the radius for C1. Which also means that the two shown circles are the same size. If that’s the case then I understand that it works. Can someone explain why it works.
Yes, the distance from Q to P is R (the radius of the known circle).
I’ll try to explain why it works but without knowing what you’re having trouble understanding I may emphasize the wrong parts. Also some of the following may be super obvious but I’m including it for clarity and not to talk down to anyone.
First any circle centered on L’ and passing through P will be tangent to L.
Secondly given a circle of radius R centered on C1 and an another point G, a circle centered on G with a radius of R plus the distance from G to C1 will be tangent to the first circle and the tangent point will be the point on the given circle that is farthest from G.
If G lies on L’ and R plus the distance from G to C1 is the same as the distance from G to P we can draw one circle tangent to both our original circle and to L (I’m pretty sure this is our goal).
We find this point by first finding Q on L’ such that Q is distance R from P (using our second radius R circle). Now all we need is to place G on L’ equidistant from C1 and Q. We accomplish this using the perpendicular bisector of QC1.
Now the distance from G to the farthest point of the original circle (the point you had labeled I) is R plus the distance from G to C1 and the distance from G to P is also R plus the distance from G to C1 since G to Q is as long as G to C1 and Q to P is R.
I hope this clears things up, but I obviously like talking about this stuff so feel free to ask questions if anything is even slightly unclear.
