Thursday, June 1, 2017

Newton's Temperature Scale

  The thermoscope, a bulb containing air with a long tube that was immersed in water, was developed by Galileo and others to measure temperature during the first half of the 17th Century. Boyle studied similar "weather-glasses" and introduced the hermetically sealed thermometer in England by 1665. In 1701 Newton anonymously published an article, Scala graduum caloris, which described a temperature scale ranging from the freezing point of water to that of a fire hot enough to make iron glow. An English translation of Newton's article can be found in Magie, A Source Book in Physics, p. 225.

Newton's temperature scale has a geometric series and an arithmetic series associated with it. The geometric series corresponds to the temperatures and the arithmetic series is associated with cooling times.

 "This table was constructed by the help of a thermometer and of heated iron. With the thermometer I found the measure of all the heats up to that at which lead melts and by the hot iron I found the measure of the other heats. For the heat which the hot iron communicates in a given time to cold bodies which are near it, that is, the heat which the iron loses in a given time, is proportional to the whole heat of the iron. And so, if the times of cooling are taken equal, the heats will be in a geometrical progression and consequently can easily be found with a table of logarithms."

After finding a number of temperatures with the aid of a thermometer, Newton describes how the hot iron was used.

"...I heated a large enough block of iron until it was glowing and taking it from the fire with a forceps while it was glowing I placed it at once in a cold place where the wind was constantly blowing; and placing on it little pieces of various metals and other liquefiable bodies, I noted the times of cooling until all these bodies lost their fluidity and hardened, and until the heat of the iron became equal to the heat of the human body. Then by assuming that the excess of the heat of the iron and of the hardening bodies above the heat of the atmosphere, found by the thermometer, were in geometrical progression when the times were in arithmetical progression, all heats were determined."

Newton's temperature scale can be constructed mathematically as follows where I've noted some corresponding temperatures on the Fahrenheit temperature scale for comparison.

The temperature point between the melting point of wax and the boiling point of water is an average. I used the geometric average which works best. One can put together a table as follows to compare the Fahrenheit temperatures with the index number, k, above.

A graphical comparison shows that the logs are fairly linear. Using 66°F for the temperature difference gave the best fit for human body temperature at the lower left of the plot.

The slope of the fitted line can be used to convert Farenheit temperatures to points on Newton's scale.

Newton's law of cooling can be in be expressed as the difference between the temperature of an object at some time and the ambient temperature being proportional to an exponential term involving time. This can to shown to be equivalent to the differential form of the law.

Supplemental (Jun 1): Leurechon Thermometer (1627)

Supplemental (Jun 2): 65°F gives a better fit for body temperature. Was this the ambient temperature at which the experiments were done? It's doubtful there was a standard temperature yet in Newton's time. For more on the history of early thermometers see Bolton, Evolution of the Thermometer, 1592-1743.

Supplemental (Jun 2): The average of the freezing point of water and body temperature is (32+98.6)/2= 65.3. Did this originate with Accademia del Cimento?

Supplemental (Jun 4): Corrected conversion formula for k.

Monday, May 22, 2017

Fermat's Problem in Three Dimensions

  Verified the Newton's method works in three dimensions. I choose the  four vertices of a tetrahedron as the given points. The Fermat point which makes the sum of the distances from the given point a minimum turned out to be mean of the given points.

I used Excel to create an anaglyph. You will need red-cyan glasses to view it properly.

Saturday, May 20, 2017

Is the Minimum for the Four Point Fermat Problem Where We Thought?

  I've been trying to convince myself that minimum in the four point Fermat problem of the last blog is not slightly displaced from the point c. The plot below shows changes in the sum L=Σℓi of the lengths of the links from the known points to the unknown point x for changes along two lines, u and v through the point (0.700,0.700) in the plane of x.

Below 0.700 both lines decrease and increase above this value. The slopes are fairly linear on each side. But it's difficult to be certain that point c is the actual minimum just going by the data because of the discontinuity in the slope. Notice that the angle from horizontal is not the same for both lines. It may be possible for the slope on the right to be decrease also but at a lower rate. But under the circumstances it does look like c is the actual point of intersection for the two line segments.

Supplemental (May 21): Obviously we can't use an extension of Newton's method to solve this type of minimum problem since the gradients are not zero at the minimum. Fermat's theory of maxima and minima is not a general theory. When doing searches for curve fits one often encounters local minima that appear to be line segments. This might happen if the minimum is paraboloidal in shape and the contour lines are elliptical.

Friday, May 19, 2017

An Insoluble Fermat Problem for the Method

 There's a four point Fermat problem that can't be solved by linearizing the function for the sum of the distances of the unknown point.

If one tries one ends up with division by zero. The gradient of L at point c is not continuous.

The distance function near a point is cone shaped.

The individual gradients are not well behaved near a given point as this plot shows.

For a problem like this one can compute the gradient function for two points displaced from the minimum and try to find where two lines in their directions through the chosen points intersect to get a better estimate of the minimum.

Thursday, May 18, 2017

An Oversight on the Fermat Point Solution

  I just noticed an error in my the solution for the Fermat Point in the last blog but it didn't affect the results. One can solve the f correction equations for dx directly and the normal equations are not needed.

One can read f|x≠0 as "f evaluated at x is not equal to zero." The normal equations are useful when one has more equations than unknowns which often occurs when one is doing least squares fits. This method might be considered the equivalent of Newton's method for finding the zero of an equation in higher dimensions.

Wednesday, May 17, 2017

Finding the Fermat Point Given Three Arbitrary Points

  At the end of a letter to Mersenne in about 1640 concerned with finding maxima and minima Fermat proposed this problem:

  "Datis tribus punctis, quartum reperire, a quo si ducantur tres rectæ ad data puncta, summa trium harum rectarum sit minima quantitas."

  "Given three points, the forth to be found, from which you draw three lines to the given points, the sum of these three lines is to be a minimum quantity."

So, given three arbitrary points, and using the method in the previous blogs, we can find the Fermat point as follows. The distances are the ℓi whose sum is to be minimized. Taking the derivative we find for an assumed value of x that dL=fTdx where f is the sum of three unit vectors pointing to x. For the position of x for the minimum value of L the change dL has to be zero for arbitrary changes in position, dx, and the only way that this can happen is if f is equal to zero too. But the value of f at the assumed point is not necessarily zero so we look at changes in f with position and see find the value of dx for which f+df=f+Mdx=0. These are the correction equations for f. The matrix M is found by extracting the derivative of the vector function f(x). Using the method of least squares one can show that dx is a solution of the normal equations.

Using the above equations in Excel and repeatedly correcting the value for x we arrive at the Fermat point after just a few iterations.

Checking the angles between the lines from x to the given points we find they are all 120° which was deduced from the minimum condition.

The correction equations for Fermat's problem are simpler than the reflection problem since we do not have a constraint on the change for dx.

Tuesday, May 16, 2017

Reflection as an Example of the Shortest Path for Light

  There's a simpler version of the Steiner Tree Problem and that is Hero's problem of finding the shortest path for a reflected ray of light. Again, for the general problem, we have the "gradient" equal to the sum of two unit vectors pointing to the unknown point, x. An additional complication is the constraint of the motion of x along a line so that dx=îdx'.

A solution for the reduced normal equations verifies that the angle of incidence equals the angle of reflection.

Reflecting the second point above the line illustrates Euclid's Prop. XX in his Elements Bk 1 asserting the sum of any two sides of a triangle is greater than the third or, equivalently, a straight line is the shortest distance between two points.

One can see that the triangles in the two problems are similar and the math works out the same.