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Interpolating


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As we are preparing to begin operations, I've been trying to wrap my head around the formula used to interpolate data on the gauging tables. We have calibrated equipment with the correction factors, but I am not 100% on interpolating corrected proofs at fractional temps above 60F.

I did find the TTB instructions ("Interpolation of Proof from Table 1 of the Gauging Manual," on TTB.gov), but am not getting what the overall formula is for the procedure, particularly as It pertains to step four of the example provided. 

Basically, if I plug in values based on corrected proofs and temps, I can get the right answer, but I can't understand why the answer is what it is. Any tips? If there's a formula out there we could use, that would be very helpful.  

 

Thanks!

 

 

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There are a couple of assumptions behind the TTB method that are not explicitly stated. In practice these assumptions have no effect on the final result, but knowing that they are there does make it easier to understand what is being done.

The first is that the interpolation that is being done should strictly be referred to as LINEAR interpolation. If we drew graphs of all the values in Table 1 we would see that the lines are curved. But in the same way that you can build a gently curving wall with straight sided bricks, you can make up a curve with lots of straight elements if the straight sections are all short enough. Because Table 1 is quite fine grained with 1 proof and 1 degree F steps we can safely use linear interpolation.

The second assumption is that the corrections for proof and temperature are independent of each other.  This is what allows the TTB method to create two separate correction factors (the 0.32 and the -0.144 in their example) and then just add them together. Again, this is not really true, but it only impacts the result in the second or third decimal place and we are not often interested in that.

I find it easier to follow these things if I can see pictures. When I studied math our calculus professor told us that at school we had been under the false impression that geometry was a branch of mathematics. Subsequently I found out that the professor was wrong and that great scientists like Paul Dirac and Albert Einstein were able to formulate their mental experiments because of their very strong understanding of geometry. So now I don't feel so bad relying on pictures and graphs.

There are two graphs attached. These follow the TTB method of treating the effects of the proof and temperature separately. I apologise to the Americans for my notation - I habitually use commas to mark the decimal but I think you use a full stop.

In the first graph it is assumed that the Indicated Proof is 80 and that we are trying to find the true proof if the temperature were 68.36F. From Table 1 we know that (when the indicated proof is 80) for a temperature of 68F the True Proof is 76.4. This is marked as point A. Similarly at a temperature of 69F the True Proof is 76.0. This is marked as point B. We join points A and B with a straight line (remember we are using LINEAR interpolation).

Now we can mark the desired temperature of 68.36 on the temperature axis. This would be 36% of the distance between the two marked temperatures. Since we have moved 36% of the distance along the temperature axis we must also move 36% of the distance on the True Proof axis. This is because we are using linear interpolation. The step on the True Proof axis is 0.4 (i.e. 76.4 - 76.0) so the vertical step is 0.36 x 0.4 = 0.144. And because the direction is downwards the 0.4 is negative and the value is taken as -0.144.

We can do a similar exercise for when the Indicated Temperature is held at 68F and we are investigating the effect of a change in Indicated Proof. In the second graph the straight line we are using for interpolation goes upwards so we get a positive value of 0.32.

Now we can combine these two effects. We know from Table 1 that at an Indicated Proof of 80 and an Indicated Temperature of 68F the True Proof is 76.4. This is our base point. The deviation from the base is -0.144 for the change in temperature from 68F to 68.36F and the deviation from the base is 0.32 for the change in Indicated Proof from 80 to 80.32. Because we have asumed these two effects are independent of each other they can just be added together to give the 76.4 + 0.32 - 0.144 = 76.576 in the TTB example.

Plugging all these values into the AlcoDens program gives a True Proof of 76.57. AlcoDens uses the full OIML equation and does not rely on linear interpolation, so the fact that the answers are so similar indicates that the underlying assumptions have not influenced the result at all.

Graph 1.png

Graph 2.png

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