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Petroleum engineering maths help

  1. Jul 12, 2011 #1


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    Hi guys. Im having trouble with my petroleum engineering course. I have attached the example thats giving me problems, start at 1.1.1 limits and just read through to the second pic where it states that the graph at 3 sec shows a height of -96.522ft and at 2 and 4 secs the height is at -48.261ft and -241.305. Am I missing something here? I have no clue where these numbers came from. Theres a formula on pic 1 but I can't seem to get the same answers and also from the graph the heights are not coinciding with time. This is a renowned handbook that pro's use so it's not a mistake. Any help appreciated!

    Attached Files:

  2. jcsd
  3. Jul 12, 2011 #2
    Well, I'm going to bet that, even though it is a renowned handbook, there's a mistake. I don't see where the -96 is coming from. It looks more like -150. The -48.261 and -241.305 coming from reading where the tangent line intersects the t = 2 and t = 4 lines, but the six-figure accuracy is ridiculous. You'd be hard-pressed to get two figures from these graphs.
  4. Jul 13, 2011 #3
    I think the only mistake is in the sentence:

    In Fig. 1.1 at the time of 3 seconds, the distance is -96.522 ft

    If should really state:

    In Fig. 1.1 at the time of 3 seconds, the slope is -96.522 ft/s

    Then, they continue to state how the can calculate the slope of the curve at 3 seconds by using the 2- and 4-second points and doing (y2-y1)/(x2-x1) ...but these x's and y's are those of the straight line, not the curve.

    As you know, the curve is represented by y = - gt2/2

    And the slope of the curve at any time is simply its derivative, and so, the equation of the straight line is y = -2gt/2 = -gt

    And so, having this last equation and substituting gravity, y = -32.174t and so, it is not ridiculous to say that at 3 seconds, y = -96.522 ...nobody said that they read the number from the graph...it came from the formula.
  5. Jul 13, 2011 #4


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    First of all, you're awesome. Secondly, shouldn't dy/dt=-32.174t as apposed to y? so actually velocity=-96.522ft/s and not distance?
  6. Jul 13, 2011 #5
    Ah! Yes, I'm sure you're right.

    Here you lose me. First, the equation of a line is y = m t + y0. If the slope of the line is -gt, then at t=3 m = -3g, and the equation of the tangent is y = (-3g)t + y0. Second, you can't assume y0 = 0. It is obviously about 48.

    Oh, come on. y at 3 sec is not -96.522, or anything close to it. Whether it came from the formula or a graph, the statement that y is -96.522 is WRONG. I think your first idea was right: they said "distance" when they meant "slope".
  7. Jul 13, 2011 #6
    To HW: You are kind of right, I am badly re-using the y...the equation is more like dy/dt = -32.174t....but for the purposes of having it in a form more familiar when ploting in the x,y cartesian plane ( more like t,y, here), I simply referred to that equation as being y = -32.174t ....we could also replace time, t, with x, if you want...and write y=-32.174x...

    To pmsrw3: There are two equations at play, here, the one for distance at any time during free falling [ y = -gt2/2] and its corresponding derivative which indicates its slope at any given time, as well [y = -gt] ....AGAIN, noticed that these two equations can be treated as two different equation and hence, I am re-using the 'y' ...but, of course, one should not forget that they are actually related...

    Anyway, for as long as HW understood...
  8. Jul 13, 2011 #7
    No, y = -gt is not the derivative equation that indicates the slope at any given time. The equation that indicates the slope at any given time is dy/dt = -gt. This is not a trivial difference.
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