Originally Posted by

**ic-racer**
Does the program you wrote for R let you solve the resulting spline equation easily for various x values; like x = 0.1 and give the resulting y value? This has been the stumbling block with the software I had been using, thus my conversion to point-to-point interpolation.

Dale, indeed you can do both ways. If you know x (say log rel E) and want to get y (say density), you just use *predict()* which takes the curve model as its first parameter, and a vector of x values for which you would like to find y—it can be just a single-valued vector, as in:

predict(my.curve, data.frame(He=1.4))

which will compute density at logE of 1.4. Since I use this a few times, I define function D that takes an x and the curve and returns the y (see around line 130 in the code). However, to do it the other way, that is to find the x for which y has a certain value, say to find the logE at which density is 0.1 over fb+f, you use *uniroot(),* which takes a function, including non-linear ones that may results from using splines, and looks for an x at which y is 0. So if you want to find x for y=0.1, you just subtract it from the value. The call would be something like:

uniroot(function (x) D(x) - 0.1, c(0:3), curve.model)

assuming D was the function mentioned earlier. Let me know if you would like me to add code for this into the script. I might extend the code, at a later point, to look for ISO triangles, and possibly the other gradient calculations.