Math can be useful in determining truth vs perception
by Ann
(Indiana)
Another way to convert Fahrenheit to Celsius
Math can be very useful when trying to make a decision that involves money, or time and distance, or building, or many other things. We can "feel" or "intuit" many mathematical answers. Sometimes our "intuitive" math is far more efficient that actual calculations -- for instance, we can easily catch a ball, but the math we must use to describe its predicted path is differential calculus (which very few of us learn).
Sometimes our instinctive impressions can lead us astray, however. When dealing with complex systems that we have little direct, and repeatedly tested, experience with (such as playing catch), mathematics will be far more reliable. With weather systems, it is unlikely that we have any direct experience with conditions in the upper atmosphere with which to "feel" our answers.
While this page can be helpful to a layman attempting to understand how meteorologists use temperature conversions, it still refers to areas of knowledge a layman has not yet obtained. The links provided might lead to a deeper understanding, but at first glance it is mostly confusing and seems to contain inconsistencies.
For instance, the part, "We choose one proportional to the
Atmosphere layer thickness..." caused me to immediately ask, "Choose "one" what? Choose a system of conversion, choose an atmospheric layer, choose a specific temperature?"
The chart further down the page seems to me to show data
different than the explanation claims. It says, "we determine that the average temperature between 550 millibars and 650 mbar (3774 metres above sea level and 5076m) was about -5 degrees Celsius...." but when I look at the two triangular areas on the chart (which have top and bottom margins at the 550 and 650 mb lines) and follow the other chart lines to the Celsius scale at the bottom, I find answers other than -5 degrees. If I go straight down, I find about 15 degrees; if I follow the red line, I find about 2 degrees; if I follow the curved yellow lines, I find a range around 40 degrees. So, I do not understand at all how to use that chart correctly.
Also, your explanation of Fahrenheit/Celsius conversion is not the way I learned it, so I had to check the math before I could accept it as working. I learned C*9/5+32 and (F-32)*5/9 - since 40*9/5-40=32, it works out, and maybe your method is less prone to error as both conversions involve adding before multiplying and dividing, so maybe those using it are less likely to forget their parenthesis, but the method I learned makes more "intuitive" sense to me.
Barry's Response - You have inspired me to make some changes, Ann. I hope it reads a little better now.
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