Shortcut, common or not always useful?
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Shortcut, common or not always useful?
I was just doing some unit conversions as my physics teacher just is getting us back into everything etc. and I noticed that if I made 1 a placeholder for dimensional units and then converted it and then put it to the power specified by the units (working with cubed units means I cube my converted answer, squared means I square it etc.) I would be able to divide whatever was converted on the other side by this to obtain a correct conversion.
So for psi to N/m^2 I converted pound-force to Newtons and then divided that by (1/39.3700787)^2 and got a correct conversion. The psi was 30, I got 206842.7 N/m^2. The Wolfram alpha value was 206843 N/m^2.
It's just that this method seemed a little unorthodox to me, so I wasn't sure if it would work in all situations, if it was a legitimate shortcut, or if it was a common way of solving this type of problem.
So for psi to N/m^2 I converted pound-force to Newtons and then divided that by (1/39.3700787)^2 and got a correct conversion. The psi was 30, I got 206842.7 N/m^2. The Wolfram alpha value was 206843 N/m^2.
It's just that this method seemed a little unorthodox to me, so I wasn't sure if it would work in all situations, if it was a legitimate shortcut, or if it was a common way of solving this type of problem.
Guywithpants- Settlers
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Join date : 2011-08-21
Age : 28
Location : Pennsylvania, United States
Re: Shortcut, common or not always useful?
I like to write it this way:
I can calculate the dimensionless scale factors from one dimension to another:
and I can use this to calculate the scale for [psi] to [pascals]:
[psi] = [lb/inch^2] = [lb/inch^2] [meter^2/Newton] [Newton/meter^2] = [lb/Newton] [meter/inch]^2 [pascals] = 4.4482.. * (39.37..)^2 [pascals] = 6894.7.. [pascals]
This is exactly what you said but written in a nice notation (actually this notation is invented by James Clerk Maxwell). I think everything you said is correct except you made arithmetic errors.
The reason this works is because the are just scale factors of others, that is, if A and B are units then x A = [B/A] x B where [B/A] is some number. On the other hand, not all units work this way: All three of Kelvin, Celsius and Fahrenheit have different zero points so they are linearly related but not just by scale factors. In the case of units like this of course you must deal with it differently.
- [psi] = [lb/inch^2]
- [pascals] = [Newton/meter^2]
- [meter] = 39.37.. [inch]
- [lb] = 4.4482.. [Newton]
I can calculate the dimensionless scale factors from one dimension to another:
- [meter/inch] = 39.37..
- [lb/Newton] = 4.4482..
and I can use this to calculate the scale for [psi] to [pascals]:
[psi] = [lb/inch^2] = [lb/inch^2] [meter^2/Newton] [Newton/meter^2] = [lb/Newton] [meter/inch]^2 [pascals] = 4.4482.. * (39.37..)^2 [pascals] = 6894.7.. [pascals]
This is exactly what you said but written in a nice notation (actually this notation is invented by James Clerk Maxwell). I think everything you said is correct except you made arithmetic errors.
The reason this works is because the are just scale factors of others, that is, if A and B are units then x A = [B/A] x B where [B/A] is some number. On the other hand, not all units work this way: All three of Kelvin, Celsius and Fahrenheit have different zero points so they are linearly related but not just by scale factors. In the case of units like this of course you must deal with it differently.
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