# In Praise of Dimensional Analysis

I suppose so, but my first thought is to convert to meters/minute**2
-- glen
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glen herrmannsfeldt wrote:

Nah, furlong-seconds per cubic fortnight.
Cheers,
Phil Hobbs
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Phil Hobbs wrote:

A furlong per fortnight is a velocity. That leaves second per square fortnight (a frequency) if we factor it out. Frequency of what?
Microlightyear per century is another interesting speed.
Jerry
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Engineering is the art of making what you want from things you can get.
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glen herrmannsfeldt wrote:

Miles/hour/second is a pretty common unit. Of course, an hour-second equals a square minute. :-)
Jerry
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Engineering is the art of making what you want from things you can get.
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Jerry Avins wrote:
(snip)

One of my least favorite units is the one power companies like to use to describe usage: Kilowatt hours/day. Kilowatt seems like a fine power unit, without an extra factor of 24.
-- glen
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glen herrmannsfeldt wrote:

I think you mean that a KW-hour is a unit of energy. It is also (indirectly) a unit of revenue. Hence the usage.
Jerry
--
Engineering is the art of making what you want from things you can get.
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Jerry Avins wrote:

Glen,
Your timestamp seems to be a bit into the future.
Jerry
--
Engineering is the art of making what you want from things you can get.
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But are the units right?
;)
Eric Jacobsen Minister of Algorithms Abineau Communications http://www.ericjacobsen.org
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Jerry Avins wrote:

No, you meant that KW-hr/day is a unit of power. My mistake.
...
Jerry
--
Engineering is the art of making what you want from things you can get.
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glen herrmannsfeldt wrote:

Unfortunate for technical people, but good for 'normal' customers.
Everyone knows what a day is, and we all get billed in KW-hours. I doubt that very many people could tell you that a KW-hour/day is a hair under 42 watts, and even fewer could do it in a blink without a calculator.
--

Tim Wescott
Wescott Design Services
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As usual Tim is right. I would like to be more explicit about it's use in control systems. It's what allows you to "close the loop" on the block diagram; for instance the plant is something like degC1/ Volt, the transducer is something like Volts/degC2 and the controller is Volts/Volts. When multiplied this give degC1/degC2 as the gain and the units to measure during verification. Of course a real diagram has a lot more terms and different units; but the loop product must balance. Inserting the open loop process variables in proper units degC1/degC2 into the equations answers many questions. One additional point is that the form is not as specialized as it seems, but carries over to generalized Differential Geometry as vector/ covector spaces. Applying dimensional analysis in this realm has allowed me to make sense of a lot of formulas. The units changes (volts->degC) correspond to mapping from one vector space to another (and back); the reduction to a gain scalar is the contraction of a covector and vector; and so forth.
RayRogers Tim Wescott wrote:

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. You raise some good points. . One of the problems with the modern unit system (Pascals, Siemens, etc) is that the fundamental meaning of the unit is lost. For example: Pressure X Area = Force. Pascals X Square meters = Pascal Meter^2. So what? You must first convert Pascal to its fundamental definition, which is Newtons/Square Meter. Now dimensional analysis makes sense: (Newtons/square meter) X (Square meter) = Newtons, which is consistent with a unit of force. . Another example is characteristic impedance of a transmission line. We've all learned that the equation is R = SQRT(L/C), where L inductance/unit length, C = capacitance/unit length. Resistance (Henries / Farads)^.5? You need to get back to the fundamental relationships among voltage, current and time for inductance, and the fundamental relationships among current, voltage and time for capacitance. v = Ldi/dt. i = C dv/dt. From these fundamental equations, you can get the fundamental units of L and C: L=VoltSeconds/Ampere, C= AmpereSeconds/ Volt. Now the equation for characteristic impedance makes sense, in terms of its fundamental units: R = SQRT([(VoltSeconds/Amps)/ (AmpsSeconds/Volt)] R = SQRT(Volts^2/Amps^2) = Volts/Amps = Ohms.
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Jon wrote:
...

So what is that in meters, kilograms, and seconds? Does it matter?
Jerry
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Jon wrote:
(snip)

Capacitance is in cm, resistance in s/cm, inductance in s**2/cm.
Capacitance per unit length is dimensionless, inductance per unit length s**2/cm**2, so sqrt(L/C) is s/cm, just like resistance!
I used to be able to do it in MKS and CGS units about equally, and sometimes Heaviside-Lorentz units. (Similar to CGS, but without so many 4pis around.)
-- glen
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glen herrmannsfeldt wrote:

That's CGS absolute, with volts in abvolts (1 abvolt = .01 microvolt) and current in abamps (1 abamp = 10 ampere) I'd rather use MKS units; so would you! :-)

Maybe you wouldn't rather. :-)
Jerry
--
Engineering is the art of making what you want from things you can get.
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Jerry Avins wrote:
(snip)

One professor, when explaining the unit system he expected in answers, explained that his house power line was 1/3 of a statvolt.
-- glen
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