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Determining depth & volume relationship with bathymetry data

  1. Jul 16, 2015 #1
    I have Bathymetry data which gives the depth-area-volume relationship of a lake. I have determined the area-volume relationship using the function Area=A*Vol^B. Now I'm not sure how to determine the similar relationship between depth and volume. Can I use the same function i.e. Depth=A*Vol^B? I want the volume as the output (as I have a time series of water depth measurements that I want to convert to volumes). If I can use the same function, is this the correct way to rearrange it in excel - Volume=EXP((LN(Depth/A))/B).

    This is probably a lot easier than I am making it, :-(
     
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  3. Jul 16, 2015 #2

    DEvens

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    Some context might help. I was unaware that lakes had depth-area-volume relationships. What is a depth-area-volume relationship?

    I'm guessing you mean: You have a series of measurements of the depth of the lake. You know the surface area as a function of the volume of the lake. And you want to convert that to volume as a function of depth.

    If that is the case then, no, the same function will not do. The volume V at depth D is the integral of the area R from depth 0 to depth D.

    ## V(D) = \int_0^D R(s) ds ##

    But you have the area as a function of the volume. So you have:

    ## dV(s)/ds = R = A V^B ##

    So you could solve this equation for ##V(s)##, noting that ##V(0)=0## is your boundary condition. (Empty lake has zero volume.)
     
  4. Jul 17, 2015 #3
    Thank you. That is very helpful! And yes you guessed right, I want the volume as a function of depth. I will be revisiting my very rusty calculus.Thanks again!
     
  5. Jul 17, 2015 #4

    HallsofIvy

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    If you assume that the lake has the same depth all around, or, equivalently, are finding the average depth, then the average depth is simply the volume divided by the surface area. I don't know where you got "Area=A*Vol^B" or even what it means.
     
  6. Jul 17, 2015 #5

    DEvens

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    It means that the lake has an area that depends on the depth it is filled, and in a very particular way.

    If you work with reservoirs behind dams then this sort of thing might be quite useful. For example, evaporation is, very approximately, proportional to area. So there is some interest in keeping the area of the lake small. However, recreation such as sailing wants the lake to have large area. And the volume in the lake is one estimate of the utility of the lake. If you are using the lake for irrigation, for example, then the volume is the ability to produce irrigation. On the other hand, if you are generating electricity you are more interested in the depth, since the column of water is what gets you pressure to drive the turbines.

    So formulas such as these would be useful in managing the fill level of a lake behind a dam.
     
  7. Jul 17, 2015 #6
    Yes, the lake does not have the same depth all round, and using the average depth would not give me the detail I need. I am looking at the effects of upstream water use on a lake with declining water levels and am therefore trying to determine the water balance of the lake under natural and altered conditions. To get a sensible estimation of lake evaporation, I need to know how the surface area of the lake changes with decreasing volume (a shallow lake will have a large surface area to volume ratio and a deep ravine type lake would have a small surface area to volume ratio). DEvens has helped me to use the "area-volume relationship" of the lake to determine the volume as a function of the depth.
     
  8. Jul 17, 2015 #7
    As I understand it you have a table of data with each row containing three numbers representing (depth, area, volume). These data will have been established by measurement of depth (Bathymetry) for each cell in a grid from which a (numerical) model of the lake has been constructed.

    It is important to realise that in this table depth refers to the depth of the lake measured from some arbitrary point (possibly the deepest point of the lake) and is not the average depth.

    Given any known parameter (depth, area or volume), simply finding the corresponding row of the table enables you to find the other two corresponding parameters (with interpolation if necessary).

    Calculus is neither necessary or helpful here.
     
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