Finding Voltage As A Function of Peak Voltage

[Meadman23]

Homework Statement

http://www.rfcafe.com/references/electrical/triangle-wave-voltage-conversion.htm

I'm trying to understand how this website figured out that V = Vpk*(2/pi)*theta

Homework Equations

Vpk*(2/pi)*theta

The Attempt at a Solution

The only thing that comes to mind is that they're using the formula for the area of triangle. This doesn't seem to be the case though because:

b = pi/2 h = Vpk

A = (Vpk*pi)/4

 

[cepheid]

The red part is a straight line. In other words, V varies linearly with θ. That means it can be expressed it the form V = mθ + b where m is the slope of the line, and b is its y-intercept. We find b by finding the value of V when θ = 0. Since the plot clearly passes through the origin (0,0), it follows that 0 = m*0 + b. Or, 0 = b.

That leaves the equation in the form V = mθ, where m is the slope. So all we need to do is find the slope of the red line. We can do that just by looking at it. Slope = rise/run. In this case, the rise is V_pk, because the line rises from V = 0 to V = V_pk. It does so over a horizontal distance of π/2, (since the line goes from θ = 0 to θ = π/2). So the run is π/2, and hence m = rise/run = V_pk/(π/2) = (2V_pk)/π. Substituting in this value for m, the equation of the line becomes:

V = mθ = [(2V_pk)/π]θ

That's all there is to it. Really simple.

 

[Meadman23]

Late response, but thanks tons for the response. I had spent HOURS trying to make sense of this; I think this is my lesson to reach out for help a little more often...