Comments on: Attention Math Wizards!! http://www.mikedidonato.com/2007/06/08/attention-math-wizards/ It's hard being a superstar Tue, 02 Dec 2008 21:01:59 +0000 http://wordpress.org/?v=2.2.2 By: mike d. http://www.mikedidonato.com/2007/06/08/attention-math-wizards/#comment-62277 mike d. Wed, 13 Jun 2007 12:51:59 +0000 http://www.mikedidonato.com/2007/06/08/attention-math-wizards/#comment-62277 Mission Accomplished! The math proof is nasty though. I had to enlist the help of one of my very wise co-workers to get through it. Mission Accomplished!

The math proof is nasty though. I had to enlist the help of one of my very wise co-workers to get through it.

]]> By: Roland http://www.mikedidonato.com/2007/06/08/attention-math-wizards/#comment-61994 Roland Sun, 10 Jun 2007 23:27:33 +0000 http://www.mikedidonato.com/2007/06/08/attention-math-wizards/#comment-61994 God, this question reminds me that I really do not miss studying maths. God, this question reminds me that I really do not miss studying maths.

]]> By: shamus http://www.mikedidonato.com/2007/06/08/attention-math-wizards/#comment-61928 shamus Fri, 08 Jun 2007 18:26:52 +0000 http://www.mikedidonato.com/2007/06/08/attention-math-wizards/#comment-61928 I've done combining and simplifing of the gibberish you gave originally and get du/dy= -U/h1-3U((h1-h2)/(h1^3+h2^3))(h1-2y) I’ve done combining and simplifing of the gibberish you gave originally and get

du/dy= -U/h1-3U((h1-h2)/(h1^3+h2^3))(h1-2y)

]]> By: mike d. http://www.mikedidonato.com/2007/06/08/attention-math-wizards/#comment-61927 mike d. Fri, 08 Jun 2007 18:10:38 +0000 http://www.mikedidonato.com/2007/06/08/attention-math-wizards/#comment-61927 by the way, the W in dFT/W = Width. by the way, the W in dFT/W = Width.

]]> By: mike d. http://www.mikedidonato.com/2007/06/08/attention-math-wizards/#comment-61926 mike d. Fri, 08 Jun 2007 18:08:01 +0000 http://www.mikedidonato.com/2007/06/08/attention-math-wizards/#comment-61926 Wow. upon review I left out a number of very useful things, without which this problem would be impossible. That's what I get for trying to only include the relevant parts of the problem. Okay. Let's try again. The goal is to find the drag imposed on a flowing fluid. drag is defined as dFT/W = - mu times the integral of (du/dy) solved from 0 - L du/dy is listed in the image above. So if we can solve for du/dy and multiply by negative mu... I'll have an answer. The problem is, that the question relates to flow under two surfaces at two different heights: h1 and h2. The length, is L. So, I have to solve the problem for L/2 first (h1) and then from L/2 to L for next (h2). X is the variable used to describe the location along underneath my surfaces. My method was this, and I'm pretty sure the method is right on. Look at surface 1 first. at h1. Substitute in my G and my Q/W. simplify once. take the integral from Zero to L/2. That adds a variable, X. solve from 0 to L/2. Simplify. NEXT in order to do the second surface, at height h2, I have to change my G equation slightly. (wow, how did I leave all this out) G becomes a function of h2 not of h1. Q/W remains the same. So I go through the same steps, this time integrating from L/2 to L. I add the two equations together... and then supposedly I'm suppose to simplify out to what was the answer in the picture above. I just can't get it though. Wow. upon review I left out a number of very useful things, without which this problem would be impossible. That’s what I get for trying to only include the relevant parts of the problem.

Okay. Let’s try again.

The goal is to find the drag imposed on a flowing fluid. drag is defined as

dFT/W = - mu times the integral of (du/dy) solved from 0 - L

du/dy is listed in the image above. So if we can solve for du/dy and multiply by negative mu… I’ll have an answer. The problem is, that the question relates to flow under two surfaces at two different heights: h1 and h2. The length, is L.

So, I have to solve the problem for L/2 first (h1) and then from L/2 to L for next (h2). X is the variable used to describe the location along underneath my surfaces.

My method was this, and I’m pretty sure the method is right on.

Look at surface 1 first. at h1.
Substitute in my G and my Q/W. simplify once.
take the integral from Zero to L/2. That adds a variable, X. solve from 0 to L/2.

Simplify.

NEXT

in order to do the second surface, at height h2, I have to change my G equation slightly. (wow, how did I leave all this out)

G becomes a function of h2 not of h1.

Q/W remains the same.

So I go through the same steps, this time integrating from L/2 to L. I add the two equations together… and then supposedly I’m suppose to simplify out to what was the answer in the picture above. I just can’t get it though.

]]> By: shamus http://www.mikedidonato.com/2007/06/08/attention-math-wizards/#comment-61925 shamus Fri, 08 Jun 2007 17:45:04 +0000 http://www.mikedidonato.com/2007/06/08/attention-math-wizards/#comment-61925 where's the x? do you mean y? where’s the x? do you mean y?

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