Formula Raytrace modifications

[omightymega]

Let me preface this post by saying that I am somewhat challanged mathmatically. I am however fairly adept at problem solving through process, but quite frankly when high school introduced me to algebra, I went to sleep.  Anyhoo......   I'm working with Ron Barnett's formula raytrace algorithm, specifically the toroidal spiral function. The params I'm specifically working on concern the xmax, xmin, xstep, and xradius, which are arrayed in the global section. I've added a param "slide" and given it a default value. Arrays in the global section read as such, and are separate arrays:
radius=@radius
xlimit=round((@xmax-@xmin)/@xstep)

 xmax determines the maximum length of the "rope" numerically, and vice versa for the minimum.
radius determines the size of the "balls" that are strung together to form the "rope"
step determines the numerical distance between the "balls" within the xmax-xmin range.

What I want to do, is tell the argument that I want the radius to be reduced, according to my @slide parameter, progressively at each increment moving away from xmax until xmin is reached.
 My attempted statement reads:  radius= radius-(@slide*x)  where "x" is an integer on a sliding scale based on the xlimit increment position as it relates to xmax and xmin.
  My chief problem is that I don't know how to tell the argument (@"x")to recognize the increment position rather than it's numerical value.

                                                  Any help?

[Nahee_Enterprises]

Let me preface this post by saying that I am somewhat challanged mathmatically.
I am however fairly adept at problem solving through process, but quite frankly
when high school introduced me to algebra, I went to sleep.

Greetings, and Welcome to this particular Forum!!   

I believe 99.9 percent of the world's population is mathematically challenged at some level or other.  The other 1/10th would never admit it.       

Have you tried discussing your problem directly with Ron Barnett??  He is usually quite helpful to those that show a real interest in working with his formulas.

[lycium]

I believe 99.9 percent of the world's population is mathematically challenged at some level or other.  The other 1/10th would never admit it.       

99.9% + 10% = 109.9% > 100%

[omightymega]

  I sent an e-mail from his "e-mail me" link on his website, but never got a response. I was probably spammed out.   Do you have a current addy for him?

[Nahee_Enterprises]

99.9% + 10% = 109.9% > 100%   

I can see quite clearly that English is not a language you understand very well !!     

First of all, I was discussing percentages as the primary subject.  The prepositional phrase "of the world's population" was only a qualifier, and not the actual numeric quantities being discussed.

So....  the "other 1/10th" was referring to a single percentage.

If you need assistance again with English, and/or with your math, please do not hesitate to speak up.       

[lycium]

I can see quite clearly that English is not a language you understand very well !!     

lol

If you need assistance again with English, and/or with your math, please do not hesitate to speak up.       

it's nice to know you deem yourself to be one in a thousand