Unwraping truncated cone shape structure

[Ingsayyad]

I want to unwrape a pillar which has cone shape. How can I do it. For cylinder I knoww CC works well.
Lower radius=1.8506 m (red in pic)
Upper radius=1.6849 m (green in pic)
Please see attached figure.

[daniel]

Well you "just" have to do some math in order to deduce the cone apex position and angle. I guess Thales is you friend here (you just need the height between the red and green circles ;).

Then the "Unroll" tool has a "Cone" mode.

EDIT: anyway, the green and red circles don't look very well "centered". Not sure what you'll get. Maybe unrolling this column as a "cylinder" would not be so bad.

[Ingsayyad]

deduce the cone apex position and angle

Can you plaese give link.

The coordinates of my pillar

Bottom circle : xyz: 22.2154,59.3347, 0
Top circle : xyz: 22.2170,59.3813,24

EDIT: anyway, the green and red circles don't look very well "centered"

Yes, The pillar is not straight.

Maybe unrolling this column as a "cylinder" would not be so bad.

It would be too much approximation. Don't you think so. I need to do for mm accuracy.

Please help me to find solution.

[daniel]

I was thinking about Thales' Intercept thorem ;)
http://en.wikipedia.org/wiki/Intercept_theorem

And for the cylinder VS cone unrolling accuracy, it all depends what you want to do with the result...

[Ingsayyad]

Dear,
I'm trying to use
unroll->cone but still not working.

my parameters are:

center:
Bottom circle : xyz: 0,0,0
Top circle : xyz: 0,0,24.6

radius:
Bottom circle : x,y,z: 1.84 m
Top circle : x,y,z: 1.68 m

Cone apex: 0,0,275 (I just calculated)
angle:0.38
projection radius=1.84 ( WHAT should I give)
I 'm getting unroll with lower and upper edge of same length.

If required, I can send you pointcoud also.

[daniel]

Oh maybe that was not clear but this is exactly what the 'unroll' function does (even for a cone). It unwrap the surface as a 'rectangle' .

This is not a 'conical' projection.

[Ingsayyad]

Should it not be like this

[daniel]

Yes it could be. But this is not what this tool is doing right now.

Maybe we could add this as an option: can you send me your cloud and the parameters so that I can update the code? (cloudcompare [at] danielgm.net)

[Ingsayyad]

surely, I will send it to you today.

Can you do me a favour to do math. I will write matlab code for my today's task.

I can calculate cylinder coordinates.
r=sqrt(x^2+y^2).............So I don't need to fix projection radius.I hope it works
theta=atan(y,x)
z=z
Now problem is what are coordinates of my projection plane.
x=?
y=? y will be fixed? it is xz-plane
z=?

[daniel]

You can look at this paper:
http://www.mamikon.com/USArticles/Rolli ... inders.pdf (page 402)

cone_unrolling.jpg (20.1 KiB) Viewed 7595 times

1) You have first to project each point on the theoretical cone (either orthogonal to the surface or radially)
2) Then you can apply the 'unroll' equation. Let A be the cone apex, alpha the opening angle and P(x,y,z) be the projected point. lets assume also that the cone axis is Z (I'll use the figure notations):

Code: Select all

s = norm(P-A)
rho = sqrt(x^2 + y^2)
phi = atan2(y,x)
theta = phi * sin(alpha)

Eventually you'll have to apply a kind of Lambert conical projection (see http://en.wikipedia.org/wiki/Lambert_co ... projection). But is should be simpler (I think theta is the longitude and the latitude is constant in your case. And you already know rho also. I don't have the time to look at the details right now ;)