Operations on Measurement UnitsDate: 08/21/98 at 13:53:06 From: Igor Subject: Measurement units operations Distance Rate = ----------- Time Why can we divide distance by time, but not add distance to time? In other words, why not to introduce a measurement unit (miles+hours) in addition to (miles/hour)? Is it possible to teach computers what is allowed to sum and what is not? I know the answer from textbooks, but I am not satisfied. Date: 08/21/98 at 18:01:20 From: Doctor Rick Subject: Re: Measurement units operations Hi, Igor. You make an interesting analogy, worthy of some exploration. You're saying that we can take the ratio of quantities with two different units, and the result is a quantity with units that combine the two: meters divided by seconds = meters per second. (We can also multiply to make a new unit: a foot-pound is the amount of work you do in lifting a 1-pound weight 1 foot.) So why can't we do this with addition? Let's think about why dividing distance by time is useful. If Mike walks 4 miles in an hour, his speed is 4 miles/1 hour = 4 miles/hour. If Sam walks 8 miles in 2 hours, his speed is 8 miles/2 hours = 4 miles/hour, also. And the fact that you get the same answer means something. If Mike and Sam walked side by side, neither one would pull ahead because they are walking at the same speed. What if we tried to do the same thing with addition? Mike walks 1 mile, then he catches the bus and rides for 4 hours. Sam walks 4 miles, then rides the bus for 1 hour. They have both gone 5 miles+hours, by your idea, but I can't think of anything that this tells us. Mike goes a lot farther, Sam takes a lot longer and gets more tired. They just have nothing in common. You know that you can add, say, miles and feet. But first you have to convert them into the same units: 1 mile + 500 feet = 5280 feet + 500 feet = 5780 feet. Can you see that you can't add miles and feet to get 501 miles+feet? It doesn't tell you anything worth knowing, since 500 miles + 1 foot would give the same answer. If you can't even add two different length measures so they mean anything, surely it's meaningless to add measures of different types. Think about it. There's a deeper question here about why multiplication is so different from addition. I'll have to think about that. - Doctor Rick, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 08/24/98 at 09:16:32 From: Doctor Peterson Subject: Re: Measurement units operations Hi, Igor. Since I saw Dr. Rick's answer to your question Friday, I've had some interesting thoughts I wanted to share with you. It's a great question to think about. The main thing I noticed is that units are inherently multiplicative. That is, when you talk about "3 miles" it means "3 times a mile." So I'll write 3 miles as: 3 * mi to make this clear. That's also why you convert units by multiplying; 3 times a mile is 3 times 5280 feet. Now, when you combine units, you are just using the commutative and associative properties of multiplication: (3 * mph) * (2 * hr) = (3 * 2) * (mph * hr) = 6 * mi If we try to add quantities with different units, watch what happens: (3 * mi) + (2 * hr) = ? There's nothing you can do with this, just as there's nothing you can do to simplify: 3 * x + 2 * y if x and y are unrelated variables. Only if they are related, say: y = 5 * x can you do anything with this. Then: 3 * x + 2 * y = 3 * x + 2 * (5 * x) = 3 * x + 10 * x = 13 * x Notice that the distributive property, not the commutative property, applies in this situation. In fact, if you did try to talk about added units like: 5 * (mi + hr) it would just expand to: 5 * mi + 5 * hr which is not the same as (3 * mi) + (2 * hr). So a unit like (mi + hr) is useless. It just doesn't behave like a unit. Another way to look at this is that when you combine two independent units, each of which gives one "dimension" of something, you don't get a single new "dimension," but a two-dimensional thing (a vector) that can't be measured by any single unit. If you multiply two units (independent or not), you get a new one-dimensional entity: an area. ^ ^ | | 2hr+ + 3mph+2hr +---+ | / | | | / |6mi| |/ | | +---+--------> +---+---------> 3mph As for teaching computers how to use units, there are several programs that do computations with units. I have one called StudyWorks that's meant for high school students. It treats units as multipliers, and automatically converts them to whatever unit you want. If I type: 3mph * 2hr = it responds with: 9.656*10^3m and if I change the m to mi, it gives: 6mi But if I type: 3mph + 2hr = it says "The units in this expression do not match." I presume it converts each value to standard metric units and finds that they are not equivalent. - Doctor Peterson, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/