# Help wanted in solving this equation

1. Nov 25, 2012

### paula1980

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 [Broken]

Any help is REALLY appreciated. Thank you for your time and help.

#### Attached Files:

• ###### help.png
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119
Last edited by a moderator: May 6, 2017
2. Nov 25, 2012

### SteamKing

Staff Emeritus
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.

3. Nov 25, 2012

### Mute

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.

4. Nov 25, 2012

### paula1980

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
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98.5 KB
Views:
100
5. Nov 25, 2012

### SteamKing

Staff Emeritus
This situation occurs frequently.

Using your spreadsheet, another value which satisfies the relation is hc = 1.585

6. Nov 25, 2012

### paula1980

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!

7. Nov 25, 2012

### Robert1986

Try something like Newton's Method. Or maybe a bisection method.

8. Nov 26, 2012

### paula1980

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.

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