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Roger Wright wrote: Why should this be so hard?
Ask the Bernoulli brothers.
A cable hanging under the force of gravity from 2 fixed points makes a catenary curve. It took the Bernoulli's a long time to discover the equation of this curve (using the calculus of variations).
Y = a*cosh(x/a)
where a = height of the cable above ground at it's lowest point.
Y(0) = a
Y(-d/2) = Y(d/2) = h
where d is the distance between poles and h is the height of the attached wire at the pole. (Each end is assumed to be at the same height on level ground).
Y(d/2) = h = a*cosh(d/(2a)) = a/2 * [exp(d/(2a)) + exp(-d/(2a))]
=> h = a/2 * [exp(d/(2a)) + exp(-d/(2a))]
solving for a is not trivial, thus the tables of empirical data.
http://www.powline.com/products/sagsec.html[^]
-Sean
----
Shag a Lizard
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Also known as the brachistochrone problem.
"I know which side I want to win regardless of how many wrongs they have to commit to achieve it." - Stan Shannon
Web - Blog - RSS - Math - LinkedIn
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Bassam Abdul-Baki wrote: Also known as the brachistochrone problem.
Related to the catenary problem, but not exactly the same. Brachistochrone problem is the problem of finding the path of shortest time between two points.
An example of the brachistochrone problem is the problem of finding the path light travels between two points (optical path length). Contrary to popular belief, light does not travel the shortest distance between two points, but the path of shortest time between two points.
-Sean
----
Shag a Lizard
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I stand corrected. Light is overrated anyway.
"I know which side I want to win regardless of how many wrongs they have to commit to achieve it." - Stan Shannon
Web - Blog - RSS - Math - LinkedIn
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Cool!
I've looked at Sagsec before, but I doubt the boss wants to buy it at that price. But it would be fun to write my own version. Thanks! I knew I could count on 3.3 million geniuses to find some practical solutions.
"...a photo album is like Life, but flat and stuck to pages." - Shog9
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If you do finish it, send a post.
Perhaps we can turn it into a cool game. :->
Any ideas? 
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Here (scroll down past the parabola stuff)[^] is an interesting derivation that solves for the height of the poles above the lowest point of the catenary in terms of the weight per unit length and the tension at the poles, which is at least a close starting point to what you are looking for (same as sag). Calculating the initial deformation (final sag) is left up to you, it changes the weight per unit length, but how much? I'm sure the solution assumes perfect inelasticity along the length of the cable, so it is a starting point at best. The derivation might give some additional clues, since it is based on the solution for the length of such a curve. Perhaps deriving the stretch due to tension first, would be a useful direction to go.
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Interesting article! Thanks!
"...a photo album is like Life, but flat and stuck to pages." - Shog9
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It's your lucky day Roger.
I have decided to grace your question with my massive intelect.
Roger Wright wrote: The cable types are well characterized, with weights per foot, modulus of elasticity, and tensile strength published.
Ok,you can go a s far as the "Sagsec" type of solution quoted elsewhere in this thread, but I believe you would be better off looking for a good Rule of Thumb, that can be applied to other areas and applications.
Ok weights per foot, modulus of elasticity and tensile strength mumbo jumbo are irrelevant. You must ascertain a maximum usable length per strand and be done with that. The other factors are only of real use when discussing the virtues of a tether for an orbitting satellite. You and your decent Injun bretheren have not ventured into this field yet I pressume.
I percieve the savings from using the best solution for this on a new power route to be substantial so working on a solution is admirable.
Ok lets say we have Pole1 at height h1 and Pole2 at height h2 and a hypoteneuse of distance dh.
What we now need to know for in the field rules of thumb are
i.) The lowest point of the power line (clue the lesser of Pole1 and Pole2 is the maximum boundary.)
ii.) The distance the lowest point is reached from either Pole.
iii.) A formula to calculate the lines height at any other distance from a Pole.
note this is easiest to do once kwowledge of i and ii are achieved.
The guts of this is that gravity on earth is generally consistent from one location to another, so a section if not all when Pole1 and Pole2 heights are equal will be symmetrical.
To calculate the new shape and dimesions of the line for the remainder of the line will use the already calculated figures to calculate the rest.
So guess what you now need to use interpolation.
Regardz
Colin J Davies
The most LinkedIn CPian (that I know of anyhow)
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I need an algorithm for combining values from a variable number of variable length categories.
For example, I might have an array of the following values with their associated categories:
value1 - category1
value2 - category1
value3 - category2
value4 - category2
value5 - category2
The result should be sets containing every possible combination of 1 value from each category:
{value1, value3}
{value1, value4}
{value1, value5}
{value2, value3}
{value2, value4}
{value2, value5}
I can't use C(n,k) because k is variable. Can anyone help? My brain has short-circuited on this one.
Thanks!
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you may try something like this
static void Main(string[] args)<br />
{<br />
int[] nItems = { 2, 3 };<br />
int[] number = new int[nItems.Length];<br />
Generate(nItems, number);<br />
}<br />
<br />
static void Generate(int[] CategoryItemCount, int[] currentCombination)<br />
{<br />
for (int i = 0; i < CategoryItemCount[0]; i++)<br />
{<br />
currentCombination[CategoryItemCount.Length - 1] = i;<br />
<br />
if (CategoryItemCount.Length != 1)<br />
{<br />
Generate(new List<int>(CategoryItemCount).GetRange(1, CategoryItemCount.Length - 1).ToArray(), currentCombination);<br />
}<br />
else<br />
{<br />
}<br />
}<br />
}<br />
regards
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Yes your test passed.
As of how to accomplish that have you ever tried Google?
Failing that try ![Badger | [badger,badger,badger,badger...]](https://codeproject.freetls.fastly.net/script/Forums/Images/badger.gif) .
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Hi Guys,
I’m creating an animation program and have built a Bezier spline motion editor. The user-controllable curve plots time(x) against value(y). So my challenge is to retrieve the 'y' value from a given 'x' on the curve. However, all the articles I've been able to find/understand only give a formula for calculating an x or y from 't' - the percentage along the given cubic bezier curve.
As 't' is more concentrated around areas which heavily curve (to make a smooth curve), it's not possible to calculate y from x. So I'm using what seems to be a recognised work around and iteratively guessing at which 't' value produces x, continually reducing the difference until I find the correct 't' which produces my required x. I then use this 't' to calculate y.
This all works perfectly well. However, this is an animation program so I need to optimize this process. My current technique requires up to 15 iterations before it finds an accurate value for 't'. I found an article online by Don Lanaster, “Some more cubic spline math BEZMATH.PS”. He mentions using the Newton-Raphelson method to reduce this process down to 3 iterations. Apparently this technique uses the slope of the curve to make the approximation of 't' more accurate. However, the formula he uses doesn't match the formula I currently have working to calculate a value on the curve. I'm sure I'm misunderstanding it!
At this point I have to confess I am a Math dummy. I can work my way around practical mathematical issues as long as I don’t have to resort to long formulas full of ancient Greek!
Here's an extract from Don's article:
--- Quote ----
Let's use a better ploy to get our approximation to close quickly. It is called the NEWTON-RAPHELSON method, but is much simpler than it sounds. Say we get an error of x - x1. x1 is the current x for our current guess. At x1, our spline curve has a slope. Find the slope.
The slope is expressed as rise/run. Now, on any triangle...
rise = run x (rise/run)
This gives us a very good improvement for our next approximation. It turns out that the "adjust for slope" method converges very rapidly. Three passes are usually good enough. If our curve has an equation of...
x = At^3 + Bt^2 + Ct + D
...its slope will be...
x' = 3At^2 +2Bt + C
And the dt/dx slope will be its inverse or 1/(3At^2 + 2Bt +C). This is easily calculated. The next guess will be...
nextguess = currentt + (curentx - x)(currentslope)
----- end quote ----
So here's the function I call iteratively with varying values for 't' and it returns 'x' for me to compare with what I want:
public float CalcBezierValue(float t, float A, float B, float C, float D)
{
t = 1 - t; // Reverse the normalised percentage
float F1 = t * t * t; // These vars used purely for visual clarity
float F2 = 3 * t * t * (1 - t);
float F3 = 3 * t * (1 - t) * (1 - t);
float F4 = (1 - t) * (1 - t) * (1 - t);
return A * F1 + B * F2 + C * F3 + D * F4;
}
As you can see the base formula which produces the return value is quite different from the one in the article (x = At^3 + Bt^2 + Ct + D). Consequently I’m failing to comprehend how to calculate the slope value or its inverse which seem to be required to reduce the number of calls to this function.
Can anyone help me?
Many thanks for your time,
Simon
-- modified at 12:01 Wednesday 13th September, 2006
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srev wrote: I’m creating an animation program and have built a Bezier spline motion editor. The user-controllable curve plots time(x) against value(y). So my challenge is to retrieve the 'y' value from a given 'x' on the curve. However, all the articles I've been able to find/understand only give a formula for calculating an x or y from 't' - the percentage along the given cubic bezier curve.
This is one of many, many reasons why I won't use a Bezier spline for camera or motion control. A) its always more difficult B) it is always more time-consuming and cpu-hogging.
I tend to use Catmull subdivision surface variants. Catmull-Rom for splines although I have my own variation of it. The original code was from a motion spline in AI Game Programming Wisdom[^] I modified it for uneven weighted distances and some other actions. If you are trying to do smooth motion cameras without a huge CPU overhead, I highly recommend the book. I use it for just about everything within, it's a great book all around, but in related to this there is a chapter on least-acceleration motion curves, PID controllers and subdivision splines, all slightly different approaches to motion control.
If you don't want the book, take a peek at these:
http://www.mvps.org/directx/articles/catmull/[^]
http://www.lighthouse3d.com/opengl/maths/index.php?catmullrom[^]
http://www.cubic.org/docs/hermite.htm[^]
_________________________
Asu no koto o ieba, tenjo de nezumi ga warau.
Talk about things of tomorrow and the mice in the ceiling laugh. (Japanese Proverb)
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Jeffry thanks for these links, they look really usefull. And Amazon has my money for the book you recommended.
Thanks again!
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Hi,
If this is your function, y = CalcBezierValue(t, A, B, C, D), the slope function may be (I have not tested it at all)
public float CalcBezierSlope(float t, float A, float B, float C, float D)
{
t = 1 - t; // Reverse the normalised percentage
float F1 = 3 * t * t; // These vars used purely for visual clarity
float F2 = 6 * t - 9 * t * t;
float F3 = 3 * (1 - t) * (1 - t) - 6 * t * (1 - t);
float F4 = -3 * (1 - t) * (1 - t);
// I am not sure about the minus, but since there's function composition t = 1 - t ...
return ( -1 ) * A * F1 + B * F2 + C * F3 + D * F4;
}
and your iteration function may look like
public float CalcBezierNextGuess(float x0, float t, float A, float B, float C, float D)
{
// I am not sure about the minus sign, may be a plus
return t - (CalcBezierValue(t,A,B,C,D) - x0) / CalcBezierSlope(t, A, B, C, D);
}
I might be completely wrong, but I have no time to test this.
Regards,
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Guys,
Thanks so much for everyone's feedback. From all the feedback on this and other forums I have the answer.
The function CalcBezierValue() computes the cubic Bézier curve as the polynomial:
x = A*t^3 + 3*B*(t^2)*(1-t) + 3*C*t*(1-t)^2 + D*(1-t)^3
To use the technique in the article I need the derivative:
x' = 3*A*t^2 + 6*B*t*(1-t) - 3*B*t^2 + 3*C*(1-t)^2 - 6*C*t*(1-t) - 3*D*(1-t)^2
The final code looks like this:
// Calc the derivative value of the curve at the specified percentage (rather than the polynomial)
public float CalcBezierDerivative(float P, float A, float B, float C, float D)
{
P = 1 - P; // Reverse the normalised percentage
float tR = 3 * A * (P * P);
float tS = 6 * B * P * (1 - P);
float tT = 3 * B * (P * P);
float tU = 3 * C * ((1 - P) * (1 - P));
float tV = 6 * C * P * (1 - P);
float tW = 3 * D * ((1 - P) * (1 - P));
return tR + tS - tT + tU - tV - tW;
}
Many thanks for all your responses.
Simon
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My friend posed the following question to me, that a friend of his was given in her math class:
Suppose in a simplified game of (American) football that the only possible amount of points one can score at a time are 3 or 7. What is the highest score that is impossible to achieve?
For example, 17 is a possible score (7 + 7 + 3) but 11 is not (there is no combination of 3's and 7's you can add together to get 11). Of course, since this is given in the context of a game, there can be no negative scores, so (7 + 7 - 3) doesn't work.
I wrote a small program to brute-force the solution, and searching up to 1000 I found the maximum impossible score is 11. However, I am not able to prove it. This problem reminds me of some things we did in Number Theory class but it's been a while and I've hence forgotten a lot of it.
Any pointers?
--
Marcus Kwok
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It's trivial to show that the score can never be 11, so you now just have to prove that all numbers greater than 11 can be scored. All numbers after 11 fall into one of 3 series:
12 + 3n + 0 = {12, 15, 18, ...}
12 + 3n + 1 = 13 + 3n = {13, 16, 19, ...}
12 + 3n + 2 = 14 + 3n = {14, 17, 20, ...}
So, now just show that 12 (4 field goals), 13 (2 field goals and a touchdown), and 14 (2 touchdowns) can be scored, then all other numbers can be scored by tacking on more field goals (3n).
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Fantastic! Thanks. Of course, now that I know the answer, it seems so obvious
--
Marcus Kwok
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