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Find coordinates for Radius Point
Posted by Discofro2 on April 4, 2015 at 12:44 pmOk, another COGO curve question
Given coordinates for PC, PT, the radius length, Delta angle and length of curve, how do I calculate a the coordinates for the radius point. I have calculated the long chord.
I am building a calculator in excel for calculating stations and offsets.
Thanks for your help.
mark-omarra replied 6 years, 1 month ago 10 Members · 15 Replies- 15 Replies
distance-distance intersection from the PC & PT is one way
if you’re using CAD, you can offset the two tangent lines into the curve the radius distance and then intersect them
> Ok, another COGO curve question
>
> Given coordinates for PC, PT, the radius length, Delta angle and length of curve, how do I calculate a the coordinates for the radius point. I have calculated the long chord.
>
> I am building a calculator in excel for calculating stations and offsets.
>
> Thanks for your help.Did you get the station and offset from coordinates routine worked out? This is just the opposite, the radius is just an offset of the PC and PT stations
Need More Info For Radius Point
Given PC, PT, Radius, Curve Length, Delta Angle, Chord Bearing and Chord Distance you still cannot determine the Radius Point, since with all the prior information there are two solutions. You need some method of indicating the curve direction, commonly one says a curve to the left or right from the tangent approaching the PC. However not all curves are tangent, some may be tangent to the line approaching the PC and leaving the PT, others to only one or the other and some are non tangent. Curves can also be described as concave to some general direction, N, NE, E, SE, S, SW, W and NW usually suffice.
You need an unambiguous way to enter curve data.
Paul in PA
station offset complete straight line complete. I’m having problems on the curve.
My plan was to calculate the inverse between the shot point and the radius point and solve from there but I’m not having trouble figuring out my RP coordinates.
Need More Info For Radius Point
Right curve … Sorry
As I Said, Curve Right or Left Does Not Always Work
It only works for a curve tangent to the approaching line.
Alternatively one could say a curve with the given chord on the right, but be careful, a curve to the right of a given chord is a curve to the left.
Paul in PA
What about a perpendicular bisector of the long chord to the radius point.
Use 1/2 delta to solve the right triangle,
then 90 degrees off the long chord azimuth back to the radius point,
use the long leg of the triangle to calculate coordinates?You have the delta…
(180-delta)/2 is the angle from the chord to the radial line.
If you are talking about road cl calculations, the easiest way to calculate coordinates on the radius point is to turn 90 degrees off the incoming or outgoing centerline bearing, at the PC or PT, and go the radial distance.
Note … that only works if the curves are tangents and, of course, would not work for a reverse curve or compound curve with no straight lines connecting to the PC or PT.
Otherwise, as others have said, if you know the PC and PT coordinates, then a distance-distance intersection works.
The solutions are the red dots, top row of circles, on web page:
http://mathworld.wolfram.com/Circle-CircleIntersection.htmlRemember that in your case R1 = R2 and that modifications need be made to account for the fact that your chord coordinates are not (0,0) and (d,0) but instead (n1,e1) and (n2,e2). Note the solutions are completely independent of the entering and exiting tangent directions.
follow-up: Find coordinates for Radius Point
As an after supper exercise, decided to derive the equations for determining the center coordinates of a circle as a function of two points on the circle. It accounts for rotation as well as allowing for neither (of the two solutions) center point at coordinates (0,0). As in the Wolfram method, this method also provides a solution independent of the entry and exit tangent directions.
Before computing, it should be verified that the chord is not a diameter. This can be done by checking that the central angle is not a straight angle or by comparing the distance between the points to twice the radius. If the chord is a diameter, then the center coordinates are the same as those for the mid point.
Determining whether the center coordinates are right or left of the curve can be accomplished by computing the area of the triangle formed by the two radii and the chord using the “area by coordinates” method. In one case, the area will be positive and the other negative. One more check is necessary. If the central angle is less than a straight angle, then the area solution method is correct. If the central angle is greater than a straight angle, then you have an opposite condition.
A check of the equations, especially the signs on terms, is in order. I did this in a hurry and did not check it. Any assistance in checking will be welcomed.
This method is one of many. There certainly are many ways to accomplish this task as evidenced by the suggestions already posted. Every method has its advantages and disadvantages. Good luck with your spread sheet.
If you don’t like distance-distance intersections with the radius, use the perpendicular bisector of the long chord and the radius for a bearing-distance intersection, or for a check anyway.
follow-up: Find coordinates for Radius Point
Dave:
In the solution I presented, the solution for h that I derived by subtracting the third equation from the fourth and rearranging terms produces the same result as that obtained by developing the equation of the line through the circle center and chord midpoint. You may notice that I incorporated f and g for convenience and that it reveals the relationship between h and k in a manner similar to the standard linear form of y=mx+b. So, even though one begins with circle-circle (distance-distance) equations, the solution leads to a line-circle (bearing-distance) intersection.An aspect I did not discuss more was the positional relationship between the chord and center point. I presented the “area by coordinates” method because most surveyors are familiar it. Determining the cross product of the vectors from one chord point to the center and from the same chord point to the other chord point yields a vector normal to the x-y plane. Dependent upon the right-hand rule, if the vector points up, positive z, then the center point is right of the chord and vice-versa. The z component of the vector is (h-x1)(y2-y1)-(k-y1)(x2-x1).
I did check my equations and found typos, although not in the solutions for h, f and g but, unless someone requests, I will not post for now.
I wouldn’t let my guys sit for the FLS test before they could work a three point curve, longhand, solving for all parts of the curve, including the radius point, beginning with only three coordinate pairs. They would tell you there are multiple ways to answer this question.
I realize this is an old post and you may not reply. But, I’ll give it a go.
Being able to determine the RP coordinate from the two point solution using distance-distance intersection given: the PC coordinate, PT coordinate, where, Radius1=Radius2, Length and Chord Bearing & Distance for non-tangent curve at both the PC and PT would be fantastic if it’s indeed possible.
Your idea to use Area by Coordinates is creative, I haven’t checked your hypothesis but, it strikes me as unlikely. I wouldn’t be surprised if the two areas calculated are equal except for the positive and negative sign. Given that the 2 radius lengths are equal.
I would like to request your solution that you presented. If the solution you presented is the link to Wolfram MathWorld in this thread being the two red dots on the top row. Then Radius1 doesn’t = Radius2 as in my case and could very well be true. But, the two red dots on the top row you refer to it appears Radius1 = Radius2. And thus would be of interest to me.
But you refer to “h derived by subtracting the third equation from the fourth”. That variable h, k, f, g, and z are not involved in the Link to the Wolfram web page.
Please explain your solution that you presented. I seemed to missing something.
Thanks in advance,
Mark o
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