This is a math problem. You want a function f(x) such that f(0) = 48, f(100) = 18.
Let's pick the easiest logical function, a linear relationship, which says f(x) can be written as f(x) = mx + b. Let's substitute in the two end conditions, and see if we can determine m and b.
f(0) = m*0 + b = 48, b = 48
f(100) = 100 m + b = 18, 100 m + 48 = 18, 100 m = -30, m = -0.3
So it looks like f(x) = -0.3x + 48 should work, that is, if we put in 0, we should get 48, and if we put in 100, we should get 18.
Let's try it with 0, 50, and 100.
f(0) = 48 (that was easy, and "correct")
f(50) = -15+48 = 33 (OK, we didn't know that before)
f(100) = -30 + 48 = 18 (also "correct").
So does f(50) make sense? 50 is halfway between 0 and 100, and 33 is halfway between 48 and 18, so maybe it does.
Now, write a LabVIEW sub-VI that takes in x (which should be between 0 and 100 -- you might want to think about adding some code to deal with numbers outside this range, either by throwing an error, "coercing" the values to this range, or whatever else you think is appropriate) and outputing the Servo position.
Bob Schor
