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Basic mathematic equation

  1. Apr 12, 2006 #1
    Hi, here is my equation of basic mathematic.. I know it looks a bit huge but actually on the paper it isn't. In the equation is one mathematic mistake, I checked several times alone and I am not able to find it, I hope i can get some help and thank you:

    If I write like this '' L1 '' (without the space between letter and number) then this means L with index 1. I mean the same for each other letter and number where there is no space between them. Datas are the following: L is 0.25, L1 is 1480, L2 is 1680, L3 is 1660, L4 is 590, F is 21000, A1 is 1661.06, A2 is 1074.665, A3 is 132.665, E1 is 110000, E2 is 110000, E3 is 210000, Z1 is 7 * 10^-6, Z2 is 30 * 10^-6, Z3 is 12 * 10^-6 and T is 63.8

    Step Number 1: L = ((A * L1) / (A1 * E1)) + Z1 * T * L1 + ((A * L2) / (A2 * E2)) + Z2 * T * L2 + ((A + F) * L4 / (A2 * E2)) + Z2 * T * L4 + ((A + F) * L3 / (A3 * E3)) + Z3 * T * L3
    Step Number 2: A ((L1 / (A1 * E1)) + Z1 * T * L1 + (L2 / (A2 * E2)) + Z2 * T * L2 + (L4 / (A2 * E2)) + Z2 * T * L4 + (L3 / (A3 * E3)) + Z3 * T * L3) + ((F * L4) / (A2 * E2)) + ((F * L3) / (A3 * E3)) = L
    Step Number 3: A = (L – ((F * L4) / (A2 * E2)) – ((F * L3) / (A3 * E3))) / (L1 / (A1 * E1)) + Z1 * T * L1 + (L2 / (A2 * E2)) + Z2 * T * L2 + (L4 / (A2 * E2)) + Z2 * T * L4 + (L3 / (A3 * E3)) + Z3 * T * L3)
    Step Number 4: A = (0.25 – ((21000 * 590) / (1074.665 * 110000)) – ((21000 * 1660) / (132.665 * 210000))) / (1480 / (1661.06 * 110000)) + 7 * 10^-6 * 63.8 * 1480 + (1680 / (1074.665 * 110000)) + 30 * 10^-6 * 63.8 * 1680 + (590 / (1074.665 * 110000)) + 30 * 10^-6 * 63.8 * 590 + (1660 / (132.665 * 210000)) + 12 * 10^-6 * 63.8 * 1660
    Step Number 5: A = (0.25 – 0.104811 – 1.25127) / ( 8.099 * 10^-6 + 0.660968 + 1.42 * 10^-5 + 3.21552 + 4.99 * 10^-6 + 1.12926 + 5.96 * 10^-5 + 1.270896)
    Step Number 6: A = (-1.106081) / (6.27673)
    Step Number 7: A = - 0.1762

    In Step Number 4, I also tried: 7 * 10^6 instead of 7 * 10^-6 and 30 * 10^6 instead of 30 * 10^-6 (in both 30 * 10^-6) and 12 * 10^6 instead of 12 * 10^-6 but I still don't get the true result (for A) – mistake is still there, I just can't find it :frown:
     
  2. jcsd
  3. Apr 12, 2006 #2

    Integral

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    I have no idea of what you are trying to do, or even what your question is.

    What are you doing?
     
  4. Apr 12, 2006 #3

    dav2008

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    I don't understand where you are having difficulties.

    If I understand this correctly it's just a basic algebraic equation, in which you are solving for A.
     
  5. Apr 12, 2006 #4
    I am trying to solve A from this equation... It is beginner's one, so I wonder where have I done the mistake because I can't find it and this is why I made a topic here.

    Yes, I meant this.
     
  6. Apr 12, 2006 #5

    Integral

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    The first step is to get your algebra correct. Putting the numbers in is meaningless without a correct starting point.
    is this your 1st step ?


    [tex] L = \frac {A L_1} {A_1 E_1} + T Z_1 L_1 + \frac {A L_2} {A_2 E_2} + T Z_2 L_2 + \frac {(A+ F)} {L_4}} {A_2 E_2} + T Z_2 L_4 + \frac {(A+ F)}{ L_3}} {A_3 E_3} + T Z_3 L_3 [/tex]

    I question the term [itex] T Z_2 L_4 [/itex] since it does not follow the pattern established by the similar terms.

    Anything you can tell us about where you got this expression will also help.

    If I read it correctly, your 2nd step does not follow. I would appreciate it if you would click on the above formula to see how to use our LaTex feature. You may also want to look in the Latex thread in the tutorial section. Using this format makes it much easier for us to help you. TIA
     
    Last edited: Apr 12, 2006
  7. Apr 12, 2006 #6

    Ouabache

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    The easiest way to check your work in problem where the math is not difficult but rather tedious; is to use a program such as MathCAD or Matlab. I suggest MathCAD because it is displayed in standard mathematical notation, not like a programming language. Check in your school's computer lab or ask in your math department. Often MathCAD is already loaded on these PCs.
     
    Last edited: Apr 12, 2006
  8. Apr 12, 2006 #7

    Integral

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    Oh boy, I am not sure that I can condone using computerised math to avoid learning simple algebra. The only way to learn algebra is to do algebra.
     
  9. Apr 13, 2006 #8

    HallsofIvy

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    [tex] L = \frac {A L_1} {A_1 E_1} + T Z_1 L_1 + \frac {A L_2} {A_2 E_2} + T Z_2 L_2 + \frac {(A+ F)} {L_4}} {A_2 E_2} + T Z_2 L_4 + \frac {(A+ F)}{ L_3}} {A_3 E_3} + T Z_3 L_3 [/tex]

    Assuming that this is correct (and Integral has a good argument that it doesn't look right) then You need to separte terms that have an "A" from those that don't:
    [tex] L = A\left(\frac {L_1} {A_1 E_1}+ \frac {L_2} {A_2 E_2}+ \frac {1} {L_4}} {A_2 E_2}+ \frac {1}{ L_3}} {A_3 E_3} \right)+ T Z_1 L_1 + T Z_2 L_2 + \frac {(F)} {L_4}} {A_2 E_2} + T Z_2 L_4 + \frac {(F)}{ L_3}} {A_3 E_3} + T Z_3 L_3 [/tex]
    Now move all that does not involve "A" to the left side (i.e. subtract it from both sides:
    [tex] L- \left(T Z_1 L_1 + T Z_2 L_2 + \frac {(F)} {L_4}} {A_2 E_2} + T Z_2 L_4 + \frac {(F)}{ L_3}} {A_3 E_3} + T Z_3 L_3\right) = A\left(\frac {L_1} {A_1 E_1}+ \frac {L_2} {A_2 E_2}+ \frac {1} {L_4}} {A_2 E_2}+ \frac {1}{ L_3}} {A_3 E_3} \right) [/tex]
    Finally solve for A by dividing both sides by that number (in parentheses) multiplying A.
     
  10. Apr 13, 2006 #9
    HallsofIvy and Integral: This is not my first step. I tried LaTeX type of writing but it keep saying to me '' LaTeX graphic is being generated. Reload this page in a moment. ''... In my first step is difference (im talking about the one which you typed Integral) in the following thing:
    near first '' (A + F) '' (from left to right) is * L4 and under it is '' A2 * E2 ''
    and near it is '' + Z2 * T * L4 '' which is the same as '' + Z2 T L4 ''
    near second '' (A + F) '' (from left to right) is * L3 and under it is '' A3 * E3 '' and near it is it '' + Z3 * T * L3 '' which is the same as '' + Z3 T L3 ''

    I got this from mechanic construction - im trying to figure out tensions in specific cuts ''at'' the summer when the weather is more hot and this means tensions inside the mechanic construction will change, also size of it might change too. But, from the mechanic vision is everything good - i haven't done any mistake but the mistake is in mathematical part of the task. Actually those Z's are temperature's extenses (marked as symbol alpha) but to avoid higher confusion, I marked them as Z's.
     
  11. Apr 13, 2006 #10

    arildno

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    Do the problem in portions!
    Assuming the expression in Step 1 is correct, we have:
    [tex] L = \frac{A * L1}{A1 * E1} + Z1 * T * L1 + \frac{A * L2}{A2 * E2} + Z2 * T * L2 + \frac{(A + F) * L4}{A2 * E2} + Z2 * T * L4 + \frac{(A + F) * L3}{A3 * E3} + Z3 * T * L3=[/tex]
    [tex]Af_{1}+t_{1}+Af_{2}+t_{2}+Af_{3}+t_{3}+t_{4}+Af_{4}+t_{5}+t_{6}\to{A}=\frac{L-\sum_{i=1}^{6}t_{i}}{f_{1}+f_{2}+f_{3}+f_{4}}[/tex]
    With the relations:
    [tex]f_{1}=\frac{L_{1}}{A_{1}E_{1}}, f_{2}=\frac{L_{2}}{A_{2}E_{2}}, f_{3}=\frac{L_{4}}{A_{2}E_{2}}, f_{4}=\frac{L_{3}}{A_{3}E_{3}}[/tex]
    And:
    [tex]t_{1}=Z_{1}TL_{1},t_{2}=Z_{2}TL_{2}, t_{3}=\frac{FL_{4}}{A_{2}E_{2}}, t_{4}=Z_{2}TL_{4}, t_{5}=\frac{FL_{3}}{A_{3}E_{3}}, t_{6}=Z_{3}TL_{3}[/tex]

    Calculate the t's and f's separately..
    Alternatively, note for example, that
    [tex]t_{1}+t_{2}+t_{4}+t_{6}=T*10^{-6}(7*L_{1}+30*L_{2}+30*L_{4}+12*L_{3})[/tex]
     
    Last edited: Apr 13, 2006
  12. Apr 13, 2006 #11
    Yes, I meant this equation.

    Arildno, I tried what you suggested twice (in case if I did a mistake in calculating first time) but A still isn't the true result. If I understood you right then I should make A = (L - T) / F where:
    T equals the sum: t1 + t2 + t3 + t4 + t5 + t6
    F equals the sum: f1 + f2 + f3 + f4

    A = (L - 7.632727) / 0.000086881
    A = - 84975.16 :frown:
     
  13. Apr 14, 2006 #12

    arildno

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    Calculate EACH of your terms SEPARATELY!
    Let each term have a line of its own; it is totally impossible to see where you might have gone wrong elsewise.
     
  14. Apr 14, 2006 #13
    Arildno I have done exsactly this - I calculated each one separately but I got:

    t1 = 0.660968
    t2 = 3.21552
    t3 = 0.104811
    t4 = 1.12926
    t5 = 1.251272
    t6 = 1.270896
    f1 = 0.000008099
    f2 = 0.000014212
    f3 = 0.00000499
    f4 = 0.00005958
     
  15. Apr 14, 2006 #14

    arildno

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    Could you please insert into each line the appropriate form of this:
    [tex]t_{1}=Z_{1}*T*L_{1}=7*10^{-6}*63.8*1480=(answer)[/tex]

    If you could do this, I'll go over the individual calculations when I have a calculator ready.
     
    Last edited: Apr 14, 2006
  16. Apr 14, 2006 #15
    t1 = Z1 * T * L1 = 7 * 10^-6 * 63.8 * 1480 = 0.660968

    t2 = Z2 * T * L2 = 30 * 10^-6 * 63.8 * 1680 = 3.21552

    t3 = (F * L4) / (A2 * E2) = 12390000 / 118213150 = 0.104811

    t4 = Z2 * T * L4 = 30 * 10^-6 * 63.8 * 590 = 1.12926

    t5 = (F * L3) / (A3 * E3) = 34860000 / 27859650 = 1.251272

    t6 = Z3 * T * L3 = 12 * 10^-6 * 63.8 * 1660 = 1.270896

    f1 = (L1) / (A1 * E1) = 1480 / 182716600 = 0.000008099

    f2 = (L2) / (A2 * E2) = 1680 / 118213150 = 0.000014212

    f3 = (L4) / (A2 * E2) = 590 / 118213150 = 0.00000499

    f4 = (L3) / (A3 * E3) = 1660 / 27859650 = 0.00005958
     
  17. Apr 16, 2006 #16

    Ouabache

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    I use MathCAD as a check, when solving long sets of interdependent equations. It is easy to make small errors (either in the algebra or in plugging in the numbers) that can carry over to the rest of the calculations. In using MathCAD as a tool (like a calculator) for checking tedious calculations, it comes down to the same idea as what alrildno and andreii are working on. Breaking down the calculation into smaller terms and checking those results (analogous to using breakpoints in debugging a program). Computerized math should not be used to circumvent learning algebra.
     
    Last edited: Apr 16, 2006
  18. Apr 22, 2006 #17

    Ouabache

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    My derivation agree's with Arildno's algebra. I too solved each term (in MathCAD) to compare with you and they agree. I also tried the final calculation and obtain an answer similar to yours A = -84969.33, so your method looks correct. The other things to double check is that you didn't make any typos in transcribing the original equations or constants.

    what is the true result?
     
    Last edited: Apr 23, 2006
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