Defining a spline with a formula

[Anonymous]

Is it possible to define a spline with a formula using constraints or another method? I need to create a model of an ogee spillway where part of it is defined by a formula and would like to do it as accurately as possible rather than using table values. The crest of the curve needs to be tangent to the horizontal and the lower end of the curve needs to be tangent to an existing 0.75:1 slope

[JDMather · Accepted solution]

You probably only need a handful of points.

I would do the hand calculations.

A common mistake I see many uses making - is thinking more points is better, when it can be easily demonstrated that defining the curves with as few points as possible creates a smoother curve.

In addition to the "smoothness" of the curve - consider the actual fabrication process.  How close do they hold the measurements?

"Accurately as possible" doesn't make logical sense comparing CAD to physical.

In a CAD program like Autodesk Inventor with formula driven sketch, "accurately as possible" is far more accurate than possible.

[Kent1Cooper]

As a starting point, you may find PolynomialFunction.lsp helpful, available here.  It uses multiple powers of X [if you ask it to], and it uses only integer exponents, but at least it includes a way of building a Spline [or optionally, a Polyline] by calculating Y values at increments in X value, by a formula involving powers of X, and to whatever degree of precision [spacing between X positions] you choose.

[Anonymous]

Good points JDMather, I asked an engineer about the construction of the curve and it will be hand-finished to less than 1/8" to ensure that the water flows over smoothly.  I'll use that tolerance as a guideline when determining what are the least amount of points I can use to accurately define the curve.

Thanks for the link to the lisp Kent, it looks interesting but as this is a one-off I'll build the table in Excel.  If we need to do this procedure frequently I'll modify the lisp to suit.

[leeminardi]

Jon_b,

I looked at generating some points using the given equation y = 0.3454x^1.85 and found I needed to do some scaling of the equation to fit the points and get the desired slope of 1:0.75 at the end.

With a delta x of 2376 (2809 – 433) and a delta y of 1710 (139750-138040) for the curve from its crest to its transition point with the straight line, the given equation does not come close to providing correct points.

x              y = 0.34548*x^1.85

0              0

200         6242

400         22503

600         47643

800         81122

1000       122581

1200       171755

1400       228433

1600       292445

1800       363645

2000       441904

2200       527114

Using Excel’s Goal seek feature I determine a fudge factor needed to scale the equation so that at x = 2376 y is equal to 1710.  The fudge factor is 0.002814 so the equation to use is:

Y = 0.002814*0.34548*x^1.85   or Y = 0.00097203*x^1.85  

This yields the following points:

x              y

0              0.00

200         17.56

400         63.31

600         134.05

800         228.24

1000       344.89

1200       483.24

1400       642.71

1600       822.81

1800       1023.14

2000       1243.33

2200       1483.07

2376       1710.00

Note that at x=2376 the slope is the desired value of 1 : 0.75

Here's a plot of th curve:

Just multiply the y coordinates by -1 to invert the shape and reposition it at y = 139750.

I hope this helps.

Lee

[Anonymous]

Thanks Lee, I think the crest geometry is more important than the tangency at the downslope but I'll pass the information on to the engineer.

[SEANT61]

Clearly this thread has been resolved.  Thought I’d make an additional comment, though, because the drawing reveals and interesting relationship between the choice of Units, and equations containing fractional exponents. 

The use of Millimeters as a base unit requires a conversion factor (1000 * (.001^1.85)), but using Meters does not.    Another reason, perhaps, to avoid using two different units for dimensioning (let’s not even get into the 1/8” tolerance).

To analyze the validity of the formula further, differentiation converts:

F(x) = 0.3454x^1.85

To

F’(x) = 0.63899x^0.85

Solving for a slope of 1/0.75 (or 1.333333) gives the equation:

1.333333 =  0.63899X^0.85

1.333333/0.63899 = X^0.85

2.086626 = X^0.85

2.086626^(1/0.85) = X

X = 2.37584 Meters