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Old 03-02-2013, 09:05 PM   #501 (permalink)
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...
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There would be a mathematical formula for these profiles.A visit to the Math Dept. at the local university might fill in that blank.Volunteers?
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Teardrop Curve -- from Wolfram MathWorld

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Old 03-02-2013, 10:34 PM   #502 (permalink)
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The lightest green looks like a pretty good match
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Old 03-03-2013, 03:33 AM   #503 (permalink)
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kach22i -- that makes pixel pushing look trivial.

I wanted to be the first to do an overlay with the new template. When I opened it up I discovered this:


The inversed box shows the 'real world' is out of square with the scanned image. So I took the liberty of rotating it the maybe 3 it needed. You can scroll it to the window edge to see how it is now.


Then I proceeded to section an Xray view similar to the black primered bus I posted above.


It's got the curved door windows I wanted to see, and a driver's seat centered between the fender wells.

There exist aftermarket parts to stuff 14" wide tires inside the stock body so narrowing the rear track by 12" is achievable. The wheelbase could be stretched as well. And the nose blunted if it's 'too round'.
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Old 03-03-2013, 10:05 AM   #504 (permalink)
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That's mathematically fascinating. Thanks!
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Old 03-03-2013, 03:35 PM   #505 (permalink)
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That's mathematically fascinating. Thanks!
Thanks. I stumbled upon it very quickly when I searched.

I think our next step would be to get and share the ability to use it. I think a spreadsheet program would work well (Microsoft Excel). I messed with it a little bit, but my math and Excel skills are rusty.
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Old 03-04-2013, 12:25 PM   #506 (permalink)
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it could be a low Cd, right?
Needs a tail cone or something.
Add a bulbous nose on the back and run it backward.
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The power needed to push an object through a fluid increases as the cube of the velocity. Mechanical friction increases as the square, so increasing speed requires progressively more power.
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Old 03-08-2013, 12:55 AM   #507 (permalink)
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Teardrop curve

Here's the best teardrop graph I've got, so far.
  • t goes from 0 to 2*pi in 1/32 increments
  • m = 1

Note: sin^m(t/2) is (sin(t/2))^m

I just noticed the original teardrop graph has the y-axis scaled smaller than the x-axis scale. In other words, 0 to 1 on the x-axis measures 1-7/8" and 0 to 1 on the y-axis measures 1-1/8". The proportions may improve if I change the y-axis.

Which I did: Middle graph.

Then I re-scaled the y-axis on the m = 2 graph, which corresponds with the yellow line in the original: 3rd graph.

... m = 3 graph, outermost green line on original graph: 4th graph.
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Last edited by sgtlethargic; 03-08-2013 at 01:10 AM..
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Old 03-17-2013, 01:20 PM   #508 (permalink)
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Why are there no comments on the Excel graphs?
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Old 03-17-2013, 05:51 PM   #509 (permalink)
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Look at that, not so much as a Thanks. At least I could fix that part.

I Saved and will be using them as tracing jpg templates in a box modeller. I am working with geodesic nets. It would be interesting to fit one to that 4th curve, but just as an exercise—not enough interior volume for the length.

I know it can be disappointing; the hive mind has attention deficit. But we're 500 posts into a Sticky thread. There's 62,000 views overall, but you probably would get better recognition in a new, dedicated thread. I'll bet people like KamperBob and aerohead would find it and comment there.

But that's all you get, comments (and Thanks). What I would hope for is seeing them used as examples by random people in other threads. Then you will know they were noticed.
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Old 03-19-2013, 08:08 PM   #510 (permalink)
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teardrop

Quote:
Originally Posted by sgtlethargic View Post
Here's the best teardrop graph I've got, so far.
  • t goes from 0 to 2*pi in 1/32 increments
  • m = 1

Note: sin^m(t/2) is (sin(t/2))^m

I just noticed the original teardrop graph has the y-axis scaled smaller than the x-axis scale. In other words, 0 to 1 on the x-axis measures 1-7/8" and 0 to 1 on the y-axis measures 1-1/8". The proportions may improve if I change the y-axis.

Which I did: Middle graph.

Then I re-scaled the y-axis on the m = 2 graph, which corresponds with the yellow line in the original: 3rd graph.

... m = 3 graph, outermost green line on original graph: 4th graph.
Didn't mean to marginalize your contribution by a mere keystroke 'Thanks.'
I'm busier than a cat covering up ---t and I'm stealing snipits of time to sneak into here at Copy-Pro to check in on posts.
What you've shared will probably end up as the Rosetta Stone for us fabricators,allowing very discreet vectors for the difficult curvilinear forms which the air likes best,and a better way to construct them.
The teardrop remains the benchmark for low drag in ground proximity.The 'laminar' forms won't work for us due to ambient air turbulence so it's important that we be able to produce the 'traditional' drops to squeeze as much mpg as we can.
A big thanks to you and all who continue to ferret out these physics-related minutia which really ice the cake!

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