I'll take a stab at it.
According to:
the angular displacement of a spiral bimetalic thermometer in degrees is given by:
angle = (360deg/pi)*(a*L/s)*(t2 - t1)
where:
angle = rotation angle of pointer in degrres a = specific thermal expansion of strip material s = thickness of the bimetal strip L = Length of the bimetallic strip t1,t2 the starting and ending temperatures
We can combine the constants 360deg, pi, a, and s into a single constant, k, and use DT for the temperature range, yielding:
angle = k*L*DT
We can solve this for L:
L = angle/(k * DT)
If we have two lengths, L1 and L2 corresponding to two DT's with the same angular displacement, then:
L2/L1 = DT1/DT2
in this case DT2 = 130F and DT1 = 90F, so the ratio is:
L2/L1 = 9/13
Now we only need to know the initial length, L1.
The formula for the radius of an equally spaced spiral with initial radius R0 and final radius R1 over n turns is:
r(q) = R0 + c*q { q is the angle around the spiral }
where c = (R1 - R0)/(n*2*pi)
We know the initial diameter and final diameter, so In this case we have:
R0 = (1/4) inch / 2 = 3.18 mm R1 = (1-1/16) inch / 2 = 13.49 mm
so that
c = 0.178 mm/radian or 1.22 x 10^-4 inches/degree
The length of the spiral can be found by an approximation formula, or by direct integration. We'll switch to metric for lengths:
L1 = INTEGRAL(0 to n*2pi; R0 + C*Q ; dq)
= R0*n*2*pi + (1/2)*c*(n*2*pi)^2
= 484.389 mm
Now, using the ratio we found previously,
L2 = (9/13)*L1
= 335.35 mm
Length change is:
DL = L2 - L1 = -149.0 mm or 14.9 cm
This corresponds to removing 11.991 radians, or 687 degrees from the outer spiral. That's 1.91 turns.
Seems like a lot to go snipping without confirmation of the method, so as usual, free advice is worth every penny, buyer beware, you get what you pay for, etc., etc. Good luck.