Parameter vs. Temperature FormulaDate: 10/28/2002 at 21:29:53 From: Carl Subject: Parameter vs. temperature formula I am trying to figure out a formula to work out a problem. The intervals that I have to work with are: Frequency vs Temperature 1.5 0.2 2.25 0.37 3.00 1.40 The spacing between the frequency had to meet the Delta T of the temperature. The problem is to determine how accurate the spacing has to be to meet the above requirements. I need to know how to derive the delta T (f) function equation. I was told that the function equation is exponential in nature and the general form is Delta T = A + Be^(Cf), Where Am, B, C are constants. Thaks for your assistance, Carl Date: 10/29/2002 at 01:37:18 From: Doctor Jeremiah Subject: Re: Parameter vs. temperature formula Hi Carl, You have three values for f and T so you can make three equations from T = A + Be^(Cf) If you do that, you can solve for the constants A, B, and C. The three equations would be: 0.20 = A + Be^(C 1.5) 0.37 = A + Be^(C 2.25) 1.40 = A + Be^(C 3) We can rearrange then to make them easier to solve like this: ln[(0.20-A)/B] = 1.5 C ln[(0.37-A)/B] = 2.25 C ln[(1.40-A)/B] = 3 C and with a bit more manipulation of the first one: ln[(0.20-A)/B] = 1.5 C 2 ln[(0.20-A)/B] = 3 C ln[(0.20-A)^2/B^2] = 3 C which gives us these three: ln[(0.20-A)^2/B^2] = 3 C ln[(0.37-A)/B] = 2.25 C ln[(1.40-A)/B] = 3 C Now we can equate the first and third: ln[(0.20-A)^2/B^2] = ln[(1.40-A)/B] (0.20-A)^2/B^2 = (1.40-A)/B (0.20-A)^2 = (1.40-A) B^2/B (0.20-A)^2 = (1.40-A) B (0.20-A)^2/(1.40-A) = B call this equation A If we manipulate the first one again: ln[(0.20-A)/B] = 1.5 C 1.5 ln[(0.20-A)/B] = 2.25 C ln[(0.20-A)^(3/2)/B^(3/2)] = 2.25 C Now we can equate the first and second: ln[(0.20-A)^(3/2)/B^(3/2)] = ln[(0.37-A)/B] (0.20-A)^(3/2)/B^(3/2) = (0.37-A)/B (0.20-A)^(3/2) = (0.37-A) B^(3/2)/B (0.20-A)^(3/2) = (0.37-A) B^(1/2) (0.20-A)^(3/2)/(0.37-A) = B^(1/2) [(0.20-A)^(3/2)/(0.37-A)]^2 = B call this equation B Now we can manipulate equation A and B: (0.20-A)^2/(1.40-A) = [(0.20-A)^(3/2)/(0.37-A)]^2 (0.20-A)^2/(1.40-A) = (0.20-A)^3/(0.37-A)^2 (0.37-A)^2 (0.20-A)^2 = (0.20-A)^3 (1.40-A) (0.37-A)^2 = (0.20-A) (1.40-A) And solving for A: (0.37-A)^2 = (0.20-A) (1.40-A) 0.37^2 - 0.74A + A^2 = 0.28 - 1.60A + A^2 0.37^2 - 0.74A = 0.28 - 1.60A 1.60A - 0.74A = 0.28 - 0.37^2 0.86A = 0.1431 A = 0.1431/0.86 A = 1431/8600 Now we can take equation A and solve for B: (0.20-A)^2/(1.40-A) = B (0.20-1431/8600)^2/(1.40-1431/8600) = B 83521/91237400 = B And now we can use the third equation to solve for C: ln[(1.40-A)/B] = 3 C ln[(10609*91237400)/(8600*83521)] = 3 C ln[9679375766/7182806] = 3 C ln[9679375766/7182806]/3 = C So these are the three constants: A = 1431/8600 B = 83521/91237400 C = ln[9679375766/7182806]/3 Represented as numbers: A = 0.16639534883720930232558139534884 B = 9.15425034032096486747759142632e-4 C = 2.4027629321651825832225861370353 Let me know if you need more details about how I got these. - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ Date: 10/30/2002 at 07:13:31 From: Carl Subject: Thank you (Parameter vs. temperature formula) Thank you very much; you were very helpful in helping me understand the problem. Thanks again. |
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