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N00b Q: Help Converting Between English and SI

  1. Jul 19, 2007 #1
    Hi guys, I need some help converting between units. I hope you can help. :)

    The candela is defined as "the luminous intensity of a light source producing single-frequency light at a frequency of 540 terahertz (THz) with a power of 1/683 watt per steradian".

    Now, one watt is equal to: 0.737562ft-lb/s, 23.73035666ft-pdl/s, and 8.850744in-lb/s.



    So, here's the question: how would you state the definition of the candela with reference to the foot-pound/second, the foot-poundal/second, or the inch-pound/second?

    My "calculation", which may be wrong hence why I'm asking (cos I really don't know!!), is:


    "the luminous intensity of a light source producing single-frequency light at a frequency of 540 terahertz (THz) with a power of:
    1/503.754846ft-lb/s per steradian"
    1/16207.83360333ft-pdl/s per steradian"
    1/6045.058152in-lb/s per steradian"




    Now, if my answers are wrong, I would really like not only the right answers, but also a reason for the right asnwers so I know for the future :D



    Cheers guys, I really hope someone can help. :) :)
     
    Last edited: Jul 19, 2007
  2. jcsd
  3. Jul 25, 2007 #2
    bump?

    No one fancy answering? :)
     
  4. Jul 25, 2007 #3

    CompuChip

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    Science Advisor
    Homework Helper

    Just multiply it.
    If one Watt is x in new units, then 1/683 Watt is x/683 in new units.
    To check the answers: ft lb/s, in-lb/s and you can do something similar for whatever "pdl" is.
     
  5. Jul 26, 2007 #4
    okay...

    Ah right. Yes indeed. Hah! It's so obvious. Sorry, I studied Linguistics and Language; maths and physics aren't my strong points :D

    Thank you very much. :D :D :D


    (So, 1/683 Watt = around 1/926.0238ft-lb/s)
     
    Last edited: Jul 26, 2007
  6. Jul 26, 2007 #5
    Y'see, I'm trying to figure out how a system analogous to SI, but based on foot-pound units, would work (in case anyone's interested).

    Yes, I know, silly and pointless, but still... I'm curious :)


    The amp is defined as "that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to 2 x 10–7 newton per metre of length."

    Now, say we have a foot-pound system equivalent called the New Electric Current Unit (or NECU for short), and the definition for it was exactly the same as the above definition, but for "1 metre" being swapped for the word "1 foot", and the word-group "2 x 10-7 newton per metre of length" being swapped for "2 x 10-7 pound-force per foot of length", then how many amps would this NECU be equal to?


    I think the answer is the following, but please someone indulge my peculiar fetish and help me out here. :)



    Basically, let me see if I understand the definition by way of rewording it.

    If we take any one metre length of the above parallel conductors which are one metre apart, the square so formed, if there was a force of 1 (one) micronewton per square metre AKA 1 micropascal, would be caused by an electric current which we say is equal to 1 (one) "ampere" flowing thru those conductors.

    Given 1 pascal = 0.020885434233150126982210708024975 pounds per square fot (psf), then one micropascal = 0.020885434233150126982210708024975 micro-psf;

    THUS,
    1 ampere is exactly 0.020885434233150126982210708024975 NECU, right?
    That is, 1 ampere is equal to 1/47.88025898033584261612967670379 NECU




    ?
     
  7. Aug 11, 2007 #6
    No one....? heh :( :P :)
     
  8. Aug 19, 2007 #7
    Hmm, I tried figuring it out another way and am completely confused now. Heh.

    If 1 "NECU" is
    "that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 foot apart in vacuum, would produce between these conductors a force equal to 2 x 10–7 pound-force per foot of length."
    then the sum I need to do is 2 N = 0.449617886lbf divided by 3.2808398950 (the number of feet per metre), right?

    This gives me 1 Amp = 0.1370435316528 NECU; 1 NECU = 7.296951471839630 Amp.



    Is this right? I'm trying to learn, but I need someone to tell me if I'm right or not.
     
  9. Aug 19, 2007 #8
    I don't really know what you are trying to do, find another way unitary measurement of the ampere? Well the ampere is defined as the amount of charge flowing per second. You can basically make whatever units you want if you have the conversions, and perform the dimensional analysis.

    http://www.chem.tamu.edu/class/fyp/mathrev/mr-da.html
     
  10. Aug 20, 2007 #9

    Really I'm just trying to grasp the thing in solid terms. Converting to another system - this hypothetical foot-pound electrical system - helps me make it more understandable in my mind. :)


    As I say, I understand how to express the definition of the ampere in foot-pound units. However, if we have a hypothetical electrical unit (heu) so it is equal to that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 foot apart in vacuum, would produce between these conductors a force equal to 2 x 10–7 pound-force per foot of length , then I am unsure how to convert that back into amperes. That is, how many "heu" is one ampere, or how many amperes is one heu? I find that trickier to figure out. Yeh, I'm not a maths or physics person naturally... :P :) :(

    Obviously, 1 ft = c. 1/3.2808', and 1 lbf = c. 4.4482 N. So then if we expressed the definition of the ampere in foot-pound units, we would use figures. HOWEVER, how would we find out how many amps a heu was, where one heu is defined as I defined it above? :):eek:
     
  11. Aug 20, 2007 #10
    Is this simplification of the definition of the ampere fair?

    The amp is the current that produces a pressure of 2 micropascals.

    Thus, given 1 micropascal is equal to c. 0.020885 micro pounds-force per square foot, or c.1/47.88 micro pounds-force per square foot, then if we define the HYPOTHETICAL ELECTRICAL UNIT exactly as follows, then 1 HYPOTHETICAL ELECTRIC UNIT should be equal to c. 47.88 amps, right?





    The definition of the HYPOTHETICAL ELECTRICAL UNIT (HEU) is:

    that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 foot apart in vacuum, would produce between these conductors a force equal to 2 x 10–7 pound-force per foot of length



    Is this right? :S
     
    Last edited: Aug 20, 2007
  12. Aug 20, 2007 #11
    So going with your NECU example you have 2x10^-7 lb/ft

    [tex] \frac{2x10^{-7} lb_f}{ft} * \frac{4.448N}{lb_f} * \frac{3.3ft}{m} = 2.9357*10^{-6} \frac{N}{m}[/tex]

    If you want how many A that is then you want how many amps give you that answer

    [tex]y(2*10^{-7} \frac{N}{m}) = 2.9357*10^{-6} \frac{N}{m}[/tex]

    Divide both side of the equation by 2*10^-7 N/m

    [tex] \frac{2.9357*10^{-6}}{2*10^{-7}} = 14.6784[/tex]

    This means that 1 amp * (factor y) = 1 NECU

    So there are 14.6784 amps in 1 NECU. Don't dismiss dimensional analysis as something trivial that you don't need to learn because that is all I used, and it obviously made your problem easy.
     
    Last edited: Aug 20, 2007
  13. Aug 21, 2007 #12
    Mindscrape, I reaaaaaaaaaaaaally appreciate the above answer and explanation. 4.448 x 3.3 = 14.6784... which is the first answer before all that other stuff to get you back to that figure.

    Anyway, I am gonna keep reading this till it sinks in. I don't quite get it yet. But then I am knackered out of my brain. I'll post back at a later date with more comments on this :D
     
  14. Aug 21, 2007 #13
    No problem. Hopefully the first equation makes perfect sense to you. It is merely rearranging quantities. If 4.448 N = 1 lbf then when you divide both sides of the equation by either 1 lbf or 4.448 N you get one. You can multiply anything by one without effecting the math, so you are basically just using relationships to get to other relationships without actually changing any physical property (i.e. not violating mathematical laws).

    The second part is simply an algebraic expression. It's like saying if I know that shoes need to have socks to go with them, then two socks are needed for each set of shoes. So (ignoring the orders of magnitude) the amps equation I wrote is like saying how many sets of shoes will give me 29 socks, which happens to go with 1 Swedish shoe? In this situation, it is like saying how many sets of amps will give me a NECU unit? So it is the definition multiplied by some factor that gives the other unit.
     
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