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Horizontal Curve Calculations
Posted by field-dog on June 19, 2021 at 3:51 amMy old SMI card for the HP 48GX can solve a horizontal curve with inputs of arc length and external distance. I don’t know how it does it, but it seems to take a little longer to solve than usual. Input: L = 418.88′, E = 47.32′. Output: ?? = 48?ø00’06”, R = 499.99′. I see relationships between the PC/PT and where the external distance bisects the arc. Also, the angles formed between the external distance and the chords of the sub-arcs.
Another one I can’t figure out, is inputs of arc length and tangent distance. Input: L = 418.88′, T =222.61′. Output: ?? = 47?ø59’30”, R = 500.09′. I believe the angle formed by the tangents at the PC/PT and the chords of the sub-arcs is ??/4.
In this example, the L, E, and T values are from an original input of ?? = 48?ø00’00” and R = 500.00′.
mathteacher replied 2 years, 10 months ago 6 Members · 22 Replies- 22 Replies
I once proved the ??/4 relationship with geometry but don’t remember how.
Many sets of two inputs theoretically define the curve but a small change in some of the parameters makes a serious change in the curve, so I chose to not include some pairs as usable pairs in my calculator for Windows.
https://www.dropbox.com/s/p6ch5d0y3ghgeut/CurveCalc.exe?dl=0
.It has been my experience, that most of the time the delta and radius are available. What if they aren’t, because you’re using some very old or special purpose survey that lacks information? That’s why I’m interested in how to calculate missing pieces of a curve like my SMI card can do. I recently discovered the following website.
Radius calculation given Arc Length and Chord length
https://onlineconversion.vbulletin.net/forum/main-forums/convert-and-calculate/13416-radius-calculation-given-arc-length-and-chord-lengthOne reply was
Y=Chord
L=Arc
R=RadiusY^2 = (2*R^2)(1-cos((180L)/(Pi*R)))
Follow Newton-Raphson method and iterate to get an accurate R(radius).
I have no idea how to use the Newton-Raphson method or how to iterate. What programming language did you use to write CurveCalc? You mentioned CurveCalc a while back, and I did download and try it.
MH- Posted by: @field-dog
It’s been too long for me to recall Newton-Raphson in detail. I think you put the functions into the form f(x)=(mess)=0, and use that with its derivative at an initial guess for x to predict a zero crossing. If things are well-behaved that prediction is closer to the real answer so you can keep improving by repetition.
https://en.wikipedia.org/wiki/Newton%27s_method
Given arc and chord, my program attempts an answer, but warns that it is very sensitive. It uses the ratio of chord to arc length to estimate delta. Some algebraic manipulation was needed to get that into the form of the inverse of sinc(x) = sin(x)/(x radians), which I calculate in a separate routine with a polynomial approximation. Then it finds radius from arc and delta.
The program was written in C++ with Visual Studio 6 (very old).
. Field Dog, Dave Lindell can show you how to do the problem. He has posted on how to solve similar problems
here on RPLS. Look for his email address and send him a message and when he gets time I am sure he will get back to you. Also MathTeacher who post here can help, its right down his line. Its not to difficult.
JOHN NOLTON
PS does your HP48GX with the SMI card give you more than 2 place accuracy? If the real delta is 48-00-00 and the radius is 500.0000 then arc length is 418.8790205+ and the Tangent would be 222.6143426+
You need to use more places in your input data to get nearer to the correct answer; as in your problem #2
To answer your question: you will have to set up the equations and use iteration (of one form or another) to solve them. Or by series expansion of the equations, Or by approximate equations, Or by tables (if they have used arc length and External distance, (your 1st problem), Or if the table used arc length and tangent distance (your 2nd problem).
Its very good you want to learn how to solve these problems by NOT using a computer/calculator.
Field Dog; I forgot to give you the value of the external distance(calculated to more places) in my above post. Using your “delta” of 48-00-00 and a radius of 500.0000 E= 47.318139255
JOHN NOLTON
Field Dog; I do not see any solution by you. I hope you have not given up on your problem. Its not really NOT that hard and you will be rewarded in the end.
First you must know geometry & Trig. One should remember several equations or look them up in a survey book. Lets work on your second problem where you give the arc distance and the Tangent distance.
Before we start let me say that this problem can be worked in a 3X3 inch square piece of paper.
Equations: #1 T/R = tangent (delta/2) and #2 S=R (delta)
T=tangent distance (given as 418.88 in your problem
“delta” is the angle at the Radius point measured from the BC to the EC
S = arc length (In most books delta might used as the lower case Greek letter theta.
using the #2 equ. above lets solve for R = s/delta (in radians)
Substitute R= s/delta(rad.) into the first equation and you get T/(s/delta(rad.)) = tan (delta/2)
The final equation you will be working with is (T/(s/delta (in rad.)) – tan(delta in deg.decimal deg.) = 0
Now you have one equation with one unknown but you have one angle in radians and one in deg.decimals degree so you must be careful. Now we can make a table (use graph paper might be the best). Now the first part might be a little tricky; where to start our first guess of delta. lets start with 40 deg.
40 deg. 222.61/(418.88/0.698131701) – 0.363970234 = +0.007045565
42 deg. 222.61/(418.88/0.733038286) – 0.383864035 = + 0.005702554
44 deg. 222.61/(418.88/0.767944871) – 0.404026226 = + 0.004091153
46 deg. 221.61/(418.88/0.802851456) – 0.424474816 = + 0.002193353
48 deg. 221.61/(418.88/0.837758041) – 0.445228685 = – 0.000009726
Note the sign change between 46 degree and 48 degree. That tells us that the delta is greater than 46 but a little less than 48 degrees. You gave us (by your HP48 & SMI card 47d 59m 30s so we will try 2 more numbers so you can see what happens.
47d 59m30s 221.61/(418.88/0.837612597) – 0.445141550 = +0.000000113
47d59m31s 221.61/(418.88/0.0.837617445)-0.445144455 = -0.000000214
Note the sigh change; the answer is between 47d59m30s and 47d59m31s
the real answer is when the right side of the equation is 0.0000000000
So your final answer for the second problem is delta is 47d 59m 30 seconds (rounded to the nearest second and the Radius is 500.08509 which will round to 500.09. This is the same answer as your HP48 with the SMI card.
If you use arc length of 418.879 and a tangent distance of 222.614 you will get a delta of 48 00 00 and a radius of 499.999976 which will round to 500.000
Hope this helps?
JOHN NOLTON
PS in computer programming they put some limit on the number of iterations. When you use Newton’s method you get your answer quicker but there can be some problems. Read a good Calculus book to find out why. Also note that I gave you a hint in one of the above post that there are other types of ways to solve problems, research on your part will lead you to some wonderful things.
Last note: I used a HP20S calculator to solve this problem.
- Posted by: @john-nolton
PS does your HP48GX with the SMI card give you more than 2 place accuracy?
I can set the output to a maximum of nine places. The external distance and middle ordinate are limited to 4 places because of the screen size.
MH - Posted by: @john-nolton
If you use arc length of 418.879 and a tangent distance of 222.614 you will get a delta of 48 00 00 and a radius of 499.999976 which will round to 500.000
Great explanation of the math! Won’t most surveys, plats and right-of-way maps show arc lengths and tangent distances to only two places after the decimal point?
MH - Posted by: @john-nolton
Field Dog; I do not see any solution by you. I hope you have not given up on your problem. Its not really NOT that hard and you will be rewarded in the end.
I have not given up. I took analytic trig and pre-calculus in college. I see this is calculus-related because you are trying to find a number that approaches zero. Eventually, I would like to be able to program something like this first in HP User-RPL (HP 48GX) and then in C++. I always insist on learning the math first, then I can push buttons with confidence!
MH @field-dog Yes; most will.
@field-dog You said in your last line “I always insist on learning the math first”
I feel the same way.
Field Dog: most calculators will use some form of Newton’s approximation to solve these type of equations, and computer programs. In computer program “Mathematica” I could type in the final equation I gave you above, give it some starting value (the seed) and in less than a millisecond it will display the answer to the number of decimal places I request. We sure have come a long way from when we used logarithms to solve this problem.
Spot on, John. I took a bit of a longer route and used sine instead of tangent. That gives these two equations:
y = sqrt(94.64*418.88/x + 2239.1824 and
y = (418.88+47.32sin(1/2x)
where y is length of the tangent and x is delta.
My TI84 (bless its little non-turbo heart) gets x = 0.83778482 radians and y = 222.61577 feet. Then Radius = 418.88/0.83778482 = 499.9852 feet.
Here’s a graph of the two equations. Note that the first one is a hyperbola (x in the denominator, there’s another piece of it in the third quadrant) and the second is a sine wave, a periodic function. They intersect many times, so there are multiple solutions. You have to be somewhat careful with your starting value to be sure that the system converges to the correct spot.
Here’s a Geogebra sketch where the center of the curve is moveable but the 47.32 distance is fixed. Moving the center in small increments and not trying to put too fine a point on it gets 499.98615 feet and 48.00149 degrees. You pros can ignore the small circle. That’s just an old math teacher demonstrating that the tangent and radius are both radii and tangents. Thus, the right angle at point E in the diagram can be verified from two different perspectives.
Great problem. It shows one of the things that mathematicians continue to pursue — maximum information from minimal inputs.
I was asked by one of our engineers to write out how to define an existing road centerline from the survey points. I found that determining the tangents and measuring from the PI to one of the shots that appears nearest to the midpoint of the curve was a pretty good way to fix the curve. Here’s my paper if you like:
If we are discussing retracement; the default for old road/highway/railroad and even subdivision jobs was that the curve was controlled by the degree. If the curve ends up being slightly off an even degree, or 1/2 a degree it was planned as that degree.
Of course like everything with surveying this won’t always work.
- Posted by: @richard-germiller
On a long curve or sharp curve, the PI may not be able to be occupied?
. Math Teacher, very nice. I have noticed in the newer Calculus books they recommend the graphing calculator
TI84. Question; is the Geogebra hard to learn?
Yes, AP Calculus is taught from 4 perspectives: Verbal, numerical, algebraic, graphic. Problems are explained, justified using numbers, solved algebraically, and solved graphically.
No, Georgebra is not particularly hard to learn. Compared to, say, MathCad, it’s a piece of cake. Less capable but capable enough.
It’s open source and downloadable, but there are copyright restrictions, primarily prohibiting using what you draw in profit-making roles.
@mathteacher Thank you for the info.
This is not meant for occupation, it was meant to take the field located centerline shots and create an alignment for the designers to have a starting point to create the new road.
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