About Measurements and Unit Conversion
Date: 02/03/2002 at 22:02:04 From: Stephanie Subject: Can you help me with my homework? Can you please help me? I don't understand measurements.
Date: 02/04/2002 at 09:40:11 From: Doctor Ian Subject: Re: Can you help me with my homework? Hi Stephanie, The idea behind measurements, as far as math is concerned, is to provide some way to associate meaning with a number. If I say something like 5 + 7 = 12 I can think of those as 'pure numbers'; or I can think of them as representing some information about the world. For example, maybe I have a couple of boards, one 5 feet long, and one 7 feet long. I want to know the length of the boards when they are placed end to end. To do that, I add the lengths: 5 feet + 7 feet = 12 feet Note that in this case, the '12' has a much different meaning than if I'm adding the lengths of two streets (5 miles + 7 miles), or two ages (5 years + 7 years), and so on. What if one of the boards is 5 feet long, and the other is 7 meters long? Then simply adding the numbers makes no sense: 5 feet + 7 meters = ? The situation gets even sillier if I try to add the length of a board (in feet) and its weight (in pounds): 5 feet + 7 pounds = ? (And in some cases, even if the units are the same, it doesn't really make sense to perform certain operations on them. For example, if I know that the temperature of one star is 5000 degrees, and the temperature of another star is 7000 degrees, it makes no sense to _add_ the temperatures of the stars to get a 'total temperature'.) So one complexity that measurement introduces into math is that you need to be careful not to mix numbers with different 'units'. This leads to the topic of unit conversion. For example, if I want to add 5 inches to 7 centimeters, I need to know that 1 inch is the same thing as 2.54 centimeters: 1 in 5 in + 7 cm = 5 in + 7 cm * ------- 2.54 cm = 5 in + (7/2.54) in Now that I know that all the measurements are in the same units, I can forget about the units until I get to the answer, where I can add them back in: = 5 + 7/2.54 = 5 + 2.76 = 7.76 inches If you're not used to this, it can look a little like magic. But do you remember learning to add fractions? To add two fractions, you have to find a common denominator, right? 1/3 + 1/5 = 5/15 + 3/15 = 8/15 Why does this work? It's because a common denominator allows you to apply the distributive property: 5/15 + 3/15 = 5*(1/15) + 3*(1/15) = (5 + 3)(1/15) = 8*(1/15) = 8/15 The same thing is true of units: 5 feet + 7 feet = 5*(1 foot) + 7*(1 foot) = (5 + 7)(1 foot) = 12*(1 foot) = 12 feet Now, writing this all out is cumbersome, which is why no one does it; but it's important to understand that the steps are there in the background, for two reasons: 1) it makes problems involving units less mysterious, if there is a coherent explanation for what's going on; and 2) if you forget the rules for manipulating units, as long as you understand that the distributive property is behind the whole thing, you have a pretty good chance of being able to figure them out from scratch. Anyway, having said all this, there emerges a general approach to solving problems involving measurements: First, make sure that all your measurements have the same units, or at least compatible units. For example, if you're measuring speeds in miles per hour, your times should all be measured in hours; if you're measuring speeds in feet per second, your times should all be measured in seconds; and so on. Second, make sure that your units combine properly. For example, if the left hand side of your equation is supposed to be a speed, but on the right hand you're dividing time by distance (i.e., you're computing something like seconds per foot), then you have a problem that needs to be fixed before continuing. Once you've converted the units and lined them up properly, go ahead and 'do the math'. I hope this helps. Please write back if any of what I've said isn't clear, or if you're still having trouble applying these ideas to the problems you're being asked to solve. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
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