One ohm sounds insignificant, but let’s see...
I don’t have the exact resistance of the 090-5025-00 coil (which is 24-1570, or gauge 24 wire 1570 turns). But flipper coils are generally about 3 to 4 Ohms.
If the coil resistance is 3.6 ohms at 20 degrees C, what is its resistance at t2= 80 degrees? Temperature coefficient of copper is .00393 per degree C.
R2 = R [1 + a(t2 - t1)]
R2 = 3.60 [1 + 0.00393(80 - 20)]
R2 = 3.6 X 1.236 = 4.45 ohms, or a .85 Ohm increase in resistance.
So, about a 1 Ohm increase.
But .85 ohm added on to 3.6 Ohms is about 24% more resistance, or 24% less current, or 24% less magnetic flux available to create the tractive force that moves the coil core.
If a 24% loss of flux sounds bad, it gets a bit worse! B is not the force of the magnetic field, just the flux. For the force we need another equation. In the formula for the force, the B term is a squared term. So any loss in flux is a much bigger loss in force.
Stolen from Wikipedia:
Force between two nearby magnetized surfaces of area A
The mechanical force (F) between two nearby magnetized surfaces can be calculated with the following equation. The equation is valid only for cases in which the effect of fringing is negligible and the volume of the air gap is much smaller than that of the magnetized material:
F = B^2 A/2 μ0 (B squared times the Area divided by 2 times mu zero)
A is the area of each surface
μ0 is the permeability of space (very close to that of air)
B is the flux density
Area and mu zero stay the same hot or cold. But B has dropped 24%. What is the effect of the 24% reduction in flux on the Force?
Let’s use some round numbers to illustrate. Let’s say you have a cold coil current of 10 amps x 1000 turns, or B = 10,000 Ampere-turns. So, B squared is 100,000,000.
Hot coil current is 10 Amps less 24% or about 7.6 Amps. B will be 7.6 Amps times 1000 turns = 7,600 Ampere turns. B squared is 57,760,000.
Going from cold to hot is going from 100,000,000 A-t to 57,760,000 A-t or 42% less force, or only 58% of your cold flipper strength.
So a 60 degree C temperature rise, creating a .85 Ohm change in resistance can create a 42% reduction in force on the coil core. This is the very large drop off that people notice.
I just wanted people to know that it’s not friction, but the resistance change that lowers the force.