Unfurl.exe

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Matt Young
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Re: Unfurl.exe

Post by Matt Young »

Thanks, perfect :D
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Re: Unfurl.exe

Post by Phill »

Hey Chaps

For anyone else, a bit of clarification - unfurl2 will run on 64 bit machines, just as a 32 bit compatable program. I haven't worked out how to get it to run at 64 bit speed yet, thats a bit above my programming level. Thanks Jose for reposting exactly what was posted 3 posts above. :D

Cheers

Phill
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Re: Unfurl.exe

Post by Matt Young »

.... I should of read back through the posts properly :oops:
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Re: Unfurl.exe

Post by sinned »

Hi,

i have a question about Unfurl. Is there a posibility to process a pointcloud whith an elliptic footprint?

Thanks
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Re: Unfurl.exe

Post by Phill »

Dennis

You sure did hurt my brain on this one. I have attached a lisp file for Autocad.. why, because I am lazy and autocad lisp has some handy built in funtions (like intersect line). Don't worry, you dont need to import the data into Autocad, it is just a vehicle for processing the data from a pts file.

Right so, the disclaimers and excuses

1. You need to have the cloud data with the major elliptical axis running in the x axis (left to right, east to west). Sort this out in Cyclone before exporting your pts file (the file needs to be a pts file)

2. The deviation (y) is calculated by drawing a line from the point to the perpendiclar on the ellipse, this is done using an approximate method .. see the bottom of http://mathpages.com/home/kmath505/kmath505.htm for the technique I used. When I checked my test data in autocad on an ellispe 36x25m with a deviation of 5m from the ellipse, I got a deviation difference of -0.00000195m at about the mid radius (45°), and a change in arc postion of around 15mm. Your deviations are probably going to be a lot smaller than this so it is probably negliable.

To do the math propertly it looks you need to do intergration and other nasty maths like that, so I have tried to keep in simple, you are getting it for free don't forget.

3. The x is calculated by using the averge of the two radius (semi major & semi minor), and projecting the intersection of the ellipse and its perpendicular out to this radius. Once again, to do this arc length properly you need to do intergration using the simpsons rule or something like that, and I am too lazy for that. Doing it this way means you can easily overlay longitude lines over the finished product.

4. The Z stays the same.
5. The origin doesn't need to be at 0,0, just type it into autocad as a comma delimeted value
6. I got lazy with the error trapping routines. If the cloud lines up perfecly with the axis' of the ellipse, it will not calculate the point so there might be 4 tiny gaps in the output data.
7. I know an egg is not ellipitcally shaped, it just seemed like a funny name for the program.

See the picture below for a diagram showing how the values are calculated
Unegg.JPG
This hasn't had extensive testing, so keep an eye on it and if it falls over let me know and I'll try to fix any problems.

If it works, you owe me a beer or two

Cheers

Phill
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Re: Unfurl.exe

Post by stevenramsey »

I think we all owe you more than a couple of beers by now Phil.

Thanks for all your support to the forum it would be a missing something without you fantasic input.
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Re: Unfurl.exe

Post by JGaspar »

Phill wrote:Hey Chaps
..
Thanks Jose for reposting exactly what was posted 3 posts above. :D

Phill

Ok, ok... Phill.. My interest to helps Matt was bigger than my time to read all posts... hahaha...
Anyway, I agree with Steven.. When you come for Brazil, let me know and I'll pay you a couple of beers! ;)
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Re: Unfurl.exe

Post by thomascoar »

If you are unfurling a bridge span what is the best methodology when using cyclone to export the original pts.

Do you want the coordinate system to make the data look like a 'smile' when viewed from the top?

Also for the centre of cylinder - is this a coordinate anywhere along the centre of the cylinder line, or do you need to work out the exact centre?
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Re: Unfurl.exe

Post by Phill »

Tom

Ok. Yes a smile from the top is correct. I guess you are talking about a hanging bridge, ie San Fran. You need to make the x axis the alignment of the bridge, the y axis your current z axis (up into the air) and the z axis going across the bridge from side to side. The centre is just a 2d point so it shouldn't matter where it is.

On a side note if you are doing the deviation on a hanging bridge, it would be in catenary (hanging chain) which is not a curve or an ellipse. Let me know if this is the case, I might be able to do something. I have dabbled in catenary calcs before.

Cheers

Phill
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Re: Unfurl.exe

Post by thomascoar »

Cheers Phill,

No it is the undersides of a bridge that i am unfurling - so should work quite easily once I get the orientation sorted..

Thanks again.

Tom
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