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Reconstruction of 3D curve from two 2D projections.....

Posted on 1998-12-15
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Last Modified: 2010-04-06
Hai....

      If  I got two 2D projections of a 3D curve in real space how to reconstruct it(3D curve).Assuming that we know the position and orientation of those two image planes on which we got projections.

      where can I get the mathematical background for this and please suggest me the web sites and referrence books I have to refer....

      Respond me quickly.It's a part of my project work...
Urgent please.......

urs
Vijay
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Question by:vijaybrd
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8 Comments
 
LVL 3

Expert Comment

by:marcjb
ID: 1113397
This will give you links to a number of pages that deal with 3d graphics.  Hope this helps,

http://www.programmersheaven.com/links/link8.htm
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LVL 13

Expert Comment

by:Epsylon
ID: 1113398
You can find the x, y and z coordinates of each vertice in the 2 2D curves:
For example x and y in curve 1 and z and y (=same as previous) in curve 2.

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Author Comment

by:vijaybrd
ID: 1113399
Thanq Epsylon.....

      Yah.... the problem is we will be given 2d points on two image planes which are taken from two different views i.e. two diffrent positions and orientaions.We will be given the position and orientation of two image planes and two sets of 2d points of the curve on two image planes.So,how to construct the 3D curve from these 2D points....

      I think U got my point....

      Try to suggest methods for the above posed problem...

      Thanq for ur answer

vijay
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LVL 10

Expert Comment

by:Jacco
ID: 1113400
If the views are just plane (XY) (YZ) and (XZ) views it is quit simple:

XY: (3,4)
YZ: (4,2)

Then the point is (3,4,2) (X,Y,Y)

If the view are more complex the task is somewhat more complex:

First of all I assume no perspective is used in the view, so a view is just a plan (which is defined by two vectors and a point).

The method is as follows:

Known of a point in a view is that it is on a line, orthogonal on the viewing plan through the point on the plane.

The orthogonal vector is calculated easily:
It is the normalized vector constructed of the points where it cuts the individual axes.

If you have constructed the lines for both views,the desired point is on the crossing of these two lines.

Repeat this process for every point of the curve and you 3D curve is constructed.

My math is a bit rusty but I hope it helped.

Regards Jacco
0
 

Author Comment

by:vijaybrd
ID: 1113401
Mr.Jacco..

      Thanq a lot for ur answer.... But it is not clear to me.My Idea is that we can construct 2D curves (best fit) on each of the image planes using the given set of points.

      From these curves we have to select feature points and match them on both the curves and we have to construct 3D curve like that....If u r talking about "Reconstruction of 3D points from two 2D projections",I have already done it.Please suggest methods for curve ( direct methods if any)..

      Thanq

vijay
0
 
LVL 10

Expert Comment

by:Jacco
ID: 1113402
Your method will not work and I'll tell you why:

Getting the matching points on the two curves is impossible, because you don't know how far on the 3D curve you are when traveling along one of the 2D curves. Why? Because this information is missing in a 2D curve....

The way I described is the only way.

If you want to construct curves from 3D points you can just use a Splines method. They are very well documented. (I like Bezier curves better but they involve helper points which can't be easily calculated from an existing curve).

Regards Jacco
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Accepted Solution

by:
stiff earned 800 total points
ID: 1113403
The math for the operations you are describing are given in:
"The VNR Concise Encyclopedia of Mathematics", by Gellert,
Kustner, Hellwich, and Kasther (editors), 1975, Van Nostrand Reinhold (publisher).
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by:Moondancer
ID: 6875743
GREETINGS!

This question was awarded, but never cleared due to the JSP-500 errors of that time.  It was "stuck" against userID -1 versus the intended expert whom you awarded.  This corrects the problem and the expert will now receive these points; points verified.

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