# Help wanted in solving this equation

by paula1980
Tags: equation, solving, wanted
 P: 5 Hello: I have been trying to solve the following equation on MathCad or Excel, however its been quite a struggle :( Can anyone please able to help me? As you can see in the equation below, I have to plot the value of hc but its located on both right and left side of the equation. All other variable (ao, f and v etc.) are constants. http://s9.postimage.org/cvr73gwel/help.png Any help is REALLY appreciated. Thank you for your time and help. Attached Thumbnails
 HW Helper Thanks P: 4,306 Instead of plotting hc, try to solve your equation by iteration. Guess a value of hc, plug it into the RHS, and see how close you are so that LHS = RHS. Alternately, you can rearrange your original expression so that RHS - LHS = 0 = y. Then you can plot a curve of hc versus y to find out the value of hc which satisfies the original equation.
 HW Helper P: 1,391 Well, there's one solution of the equation that can be found by inspection of the equation itself: ##h_c = 0##. For any others (I suspect there's only one more real one from a quick plot of x - ln|x+1|), it's possible to express the solution in terms of the Lambert-W function it looks like, but it's probably easiest to just do as SteamKing advised.
P: 5

## Help wanted in solving this equation

Thanks Steamking and Mute for your prompt responses! Do appreciate it.

I have tried following steamking's advice but still have a sense of unsurity if I did this correctly.
Please have a look at the attached Excel file, where I have tried modelling this and, if possible, provide a feedback.

I am finding it a little tricky to grasp how one can have LHS = RHS in this equation. FYI, I have attached the equation also in the spreadsheet..its presented in a slightly different manner but is same equation as before.

Thanks.
Attached Files
 P-Calculation.xls (98.5 KB, 2 views)
 HW Helper Thanks P: 4,306 This situation occurs frequently. Using your spreadsheet, another value which satisfies the relation is hc = 1.585
P: 5
 Quote by SteamKing This situation occurs frequently. Using your spreadsheet, another value which satisfies the relation is hc = 1.585
That answer is not close to correct. According to the paper I am reading it should be near 25-30 range. Again its possible I may have made a mistake somewhere but thanks for your help!
 P: 825 Try something like Newton's Method. Or maybe a bisection method.
P: 5
I found the answer...please see the attachment. I simply needed to substrate Hc-Hc...when the difference is close to zero, that value of Hc corresponds to the answer I was seeking which was around 18 (although in the paper it states around 25) but with my variables I am obtained 18.

Thanks everyone for your help particularly SteamKing.
Attached Files
 P-Calculation (Solved).xls (99.0 KB, 6 views)

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