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2 questions: dimensions and watts

  1. Jan 30, 2009 #1
    1. The problem statement, all variables and given/known data


    I am having some trouble with the two questions below:

    1) The University of Maryland's mascot is a terrapin (a snapping turtle). It is represented on campus by a bronze statue in front of the library named Testudo. Unfortunately, students rub his nose for luck on tests, and he is beginning to wear out. A sculptor decides to propose a replacement. She builds a small scale model and discovers she needs 200g of bronze. When she is finished, she finds that she can give it two coats of finishing polyurethane varnish using exactly one small can of varnish. Her final statue will be 6 times as big as her model in all dimensions. How much bronze and how much varnish will she need for the final product (if the varnish must also cover two coats on the full scale model)?

    2) A contemporary computer chip dissipates 54 watts of power on a area of 1 square centimeter due to the transistors in that area. Assuming that transistors in the next computer generations require constant power independent of their size, estimate the power that a 1 square centimeter chip will use 3 years from now. Assume “Moore's Law" which claims that the transistor count (number of transistors per unit area) doubles every 18 months.

    2. Relevant equations

    Volume, Surface Area, Moore's Law

    3. The attempt at a solution

    Here are my thought processes on the two:

    1) If the full scale model is 6 times as big in all dimensions, then that means that the amount of bronze needed should be: 6*6*6*200 = 43200g. I am not sure at all about the varnish part.

    2) I'm guessing that the power dissipated will go up by a factor of 4 whenever the transitor count doubles (I don't exactly know why, but I thought I have heard that somewhere before), so that would mean that the power dissipation in 3 years would be: 54*4*4 = 864 watts.

    Any help would be greatly appreciated! Thanks!
    Last edited: Jan 30, 2009
  2. jcsd
  3. Jan 30, 2009 #2


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    Welcome to PF!

    Hi redwing1! Welcome to PF! :smile:

    Yes, 6 times as big in all dimensions, so 6*6*6 is the correct factor. :biggrin:

    Hint: for the varnish, the amount of varnish depends on the surface area of the statue. :wink:
    Nooo … you're not reading the question properly …

    try again! :smile:
  4. Jan 30, 2009 #3
    Thanks tiny-tim for the quick reply!

    For the varnish, would the factor be 6*6 because surface area deals with units in terms of squares (i.e. centimeter squared)? So the number of cans needed would be 36?

    For the transistor problem, the only other thing I can think of right now is that I'm wrong about the quadrupling part, and that if the number of transistors doubled, then the power would double too, giving 54*2*2 = 216 watts (*2 twice because there are 2 18 month intervals in 3 years)?

    Thanks again!
  5. Jan 30, 2009 #4


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    Hi redwing1! :smile:

    Yes, 6*6 for the varnish.

    For the transistor problem, just read the question

    transistors per unit area double every 18 months …

    and the area is constant …

    just because it's an area, that doesn't mean that dimensional analysis gives you a square …

    compare the dimensions of the input and the output …

    for example, if you knew that the varnish needed (in question 1, of course :rolleyes:) was 9 times as much, and you were asked how much metal was needed, you wouldn't say "it's a volume, so it must be 9*9*9", would you? :wink:
  6. Jan 30, 2009 #5
    Thanks again for your help tiny-tim!

    I'm still stuck on the transistor question - I just don't understand what I'm doing wrong yet. I understand that the area that the transistors are on does not change with time. So in the first 18 months there are twice as many transistors as before per the same area, and after the second 18 months, there are twice as many more again (4 times as many as there were in the beginning) per the same area. If, at the beginning, the number of transistors dissipated 54 watts of power on their area, then I don't see yet why twice as many would not cause the dissipation to increase by either a factor of 2 or 4 (my original two guesses). Thanks so much for your continued help! I feel like I'm so close! :smile:
  7. Jan 30, 2009 #6


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    oh, sorry, redwing1, I misread your answer :redface:

    I thought you'd done the same as in the previous post :rolleyes:

    yes, 54*2*2 is correct … sorry to put you off! :redface:
  8. Jan 30, 2009 #7
    Thanks for all your help tiny-tim! I look forward to getting more help from you in the future; you really explain things well! :approve:
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