Equivalent Temperatures, Different Scales

Date: 05/28/2003 at 22:22:13
From: Linnie
Subject: Ratios/Percentages

The inhabitants of Xenor use two scales for measuring temperature. On 
the A scale water freezes at O degrees and boils at 80 degrees; on 
the B scale water freezes at -20 degrees and boils at 120 degrees. 
What is the equivalent on the A scale of a temperature of 29 degrees 
on the B scale?

I tried figuring out the ranges of the B scale (140), and the A scale 
(80). Then I divided 29 by 140 to figure out the percentage of 29 on 
140, which was 35 percent. Then I multiplied 0.35 by A's range (80), 
and got 28 for the equivalent on the A scale, but it turned out to be 
wrong. I think that the method I used to try to find the answer was 
wrong, but I was wondering if you knew the corrrect way to solve this 
problem.

Date: 05/29/2003 at 12:07:10
From: Doctor Ian
Subject: Re: Ratios/Percentages

Hi Linnie,

This is basically the same thing that's going on with the Celsius and
Fahrenheit scales:

     Earth                 Xenor
  ---------------      --------------
    0 C = 32 F           0 A = -20 B 

  100 C = 212 F         80 A = 120 B

If you search our archive for the keywords

  celsius fahrenheit

you'll find a number of explanations of how to convert between two
scales that are defined in this way. 

Finding the ranges is a good place to start. Note that a change of 80
degrees A corresponds to a change of 140 degrees B. That means that
each change of 1 degree A corresponds to a change of 

  140/80 = 14/8 = 7/4  

degrees B. Looking at it in the other direction, a change of 1 degree
B corresponds to a change of 

  80/140 = 4/7

degrees A. 

Does that make sense? Once you have the scale factor, the easiest way
to do conversions is to find the difference between the temperature to
be converted and the freezing point of water. That eliminates the
offset between the two scales (i.e., the fact that water freezes at
different temperatures). 

For example, a temperature of 50 degrees B is 

  50 - (-20) = 70 

degrees B above freezing. 1 degree B is the same as 4/7 degrees A, so
the temperature on the A scale must be 

  (freezing) + 70*(4/7)

  = 0 + 40

  = 40

degrees A. That's a nice check, because 70 is in the middle of the B
scale, and 40 is in the middle of the A scale.  

Going the other way, a temperature of 40 degrees A is

  40 - 0 = 40

degrees A above freezing.  1 degree A is the same as 7/4 degrees B, so
the temperature on the B scale must be 

   (freezing) + 40*(7/4) 

  = -20 + 70

  = 50

degrees B. Again, we went from the middle of one scale to the middle
of the other. 

The problem with the method you used is that when you want to scale a
measurement, you have to make sure that the measurement is relative to
zero. Instead of computing

  29 is what percent of 140?

you needed to compute

  (29 - (-20)) is what percent of (120 - (-20))?

To see why, try thinking about two more scales, X and Y, such that

                  X    Y

     Freezing   100  -50
     
     Boiling    200   50

Now consider converting 0 degrees from Y to X.  0 is 0% of 100, but
that's not what you need to know. You need to know that 0 is halfway
between -50 and 50, i.e., 

  (0 - (-50)) is 50 percent of (100 - 0). 

This is why you have to compensate for the offset _before_ you do the
scaling. And when you do the scaling, you have to make sure you start
at the beginning of the other scale. If you just compute 50% of 200,
you'll end up converting 0 degrees X to 100 degrees Y. If you think
about that for a moment, you'll see that it can't be right. 

So the basic idea is:

  1) Subtract the offset in the first system.

  2) Scale the result, using the relative sizes of degrees
     in the two systems. 

  3) Add the offset in the second system. 

An easy way to check this is to consider what happens at the freezing
points. If you start with the freezing point in one system, 
subtracting the offset gets you to zero. (Do you see why this has to
be true?) Zero always scales to zero. If you add zero to the offset of 
the other system, you'll be at the freezing point.  

Does that make sense? 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/