Pantograph

[mklotz]

I couldn't find my notes describing how to calculate the spacing of the holes on the arms of the pantograph so I had to rethink the problem. (That doesn't bother me; frequently, on the second go-around, you'll see simplifications you missed the first time.) I decided to document the process here. That way someone who wants to build one can do it himself and I'll be able to find the derivation the next time I need it. :-)

The attached diagram shows a pantograph set up for a roughly three to one enlargement. The notation is as follows:

F = Fixed point
T = Tracer point
M = Marker point
S = Slider point
P, Q = Pivot points whose locations we wish to calculate

In the computations, we'll let:

R = pantograph Ratio

If R>1, the drawing traced will be enlarged, if R<1 the final drawing will be smaller. In the diagram, R is approximately 3, so a three to one enlargement. The ratio allows us to relate FT and TM thusly...

FM = R * FT
TM = FM - FT = FT * (R - 1)

Whatever, the spacings of the pivot hole P are, they must remain constant as the angle FPT changes; as we trace, we don't move pivot points around as that angle changes. Therefore, if we can calculate them for FPT = a right angle, we'll know them for all angles. Thus, for convenience, the diagram is drawn with FPT as a right angle.

Now, FPT = 90, then SPT = 90 and, since PSQT is a parallelogram (it must be for the pantograph to work) , all its interior angles are right angles.

By design, the length of the pantograph arms are equal so FS = SM. This implies that the angles SFM and SMF are equal. Since FSM = 90, it follows that FSM = 45 = FTP and FP = PT.

Applying Pythagoras to the FPT triangle we now have:

FP^2 + PT^2 = FT^2

but FP = PT so:

2 * FP^2 = FT^2

or:

FP = FT/sqrt(2).

and

PS = FS - FP = TQ

So, once you've picked a convenient length for FT based on your maximum value of R and the chosen stick length, FS, you can find FP from FT/sqrt(2) and PS from the latter equation.

Isn't geometry wonderful?