Re: Thermometer - Today's Math Challenge

Apr 06, 2009 0 Replies


Stone cut face of outdoor thermometer has...


>
> 180° rotational degrees for 130°F temperature change.
>
> Bimetal thermometer element has...
>
> 180° rotational degrees for 90°F temperature change.
>
> Bimetal is 9-1/4 turns (labeled Truflex P675RTM2), inner diameter is
> 1/4", outer diameter is 1-1/16"
>
> How many (partial) turns to remove from outer bimetal to match
> physical requirement with bimetal movement ?:-)

I'll take a stab at it.



According to:



formatting link



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.


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