Hello Cloud Compare Community,
First, sorry for my English.
I want to use Cloud compare for a comparison of point clouds of a measured water tower.
The aim is to make statements about point density.
The selected section has about 5 million points which are distributed in the histogram. (Screen photo)
What conclusions can be made about the values on the surface density (Precise r = 0.025m) respectively Volume density? What does the horizontal axis mean?
Why are the values so high? (surface val=50141 / Volume val= 1504252) I thought there are approximately 60000 Points within radius 0.025m, but that’s wrong.
Furthermore It is possible to output an average value of the dot density?
I would be very happy about an answer
Many greetings Sinak
surface density
Re: surface density
The histogram is a representation of the number of points that have the same value (density here). Values are in fact small intervals (you can change the numbers of 'classes' - also called 'bins' sometimes - with the mouse wheel).
Here for instance you can tell that more than 40.000 points have a volume density of 150000 pts / m^3.
And the surface (resp. volume) density is big because it's expressed in square or cubical 'units', where the unit is the one in which the cloud coordinates are expressed. If it's meters here, then CC will express the volume in m^3 while the number of neighbors is counted inside a sphere of 25 mm (e.g. 10 points inside a sphere of volume = 4/3*PI*R^3 = 65416.67 mm^3 = 0.000065 m^3 --> V = 10 / 0.000065 = 152866 pts / m^3).
You should use the 'population' measurement mode (it will simply give the number of points falling inside the sphere).
Here for instance you can tell that more than 40.000 points have a volume density of 150000 pts / m^3.
And the surface (resp. volume) density is big because it's expressed in square or cubical 'units', where the unit is the one in which the cloud coordinates are expressed. If it's meters here, then CC will express the volume in m^3 while the number of neighbors is counted inside a sphere of 25 mm (e.g. 10 points inside a sphere of volume = 4/3*PI*R^3 = 65416.67 mm^3 = 0.000065 m^3 --> V = 10 / 0.000065 = 152866 pts / m^3).
You should use the 'population' measurement mode (it will simply give the number of points falling inside the sphere).
Daniel, CloudCompare admin
Re: surface density
Hi Daniel,
Thank you for the quick reply.
It helped me to understand the program calculations.
When using the method Number of neighbors my question is, In which unit the radius is entered?
I have only comprehensible results when I enter the radius in m²/cm²
My calculation:
Volume: R = (3 * V / 4 * pi) ^ 1/3;
V = 1cm³ = 0.000001m³
R= 0,0133067m³
Or
R = (A / pi²) ½; A = 1cm² = 0.0001m²
R=0,0056m²
0n the pictures you can see that the results (1cm³) are not correct?! With R=0,0133067m³ the output is 46 Points…
Where is the mistake in my mind?
Thank you for the quick reply.
It helped me to understand the program calculations.
When using the method Number of neighbors my question is, In which unit the radius is entered?
I have only comprehensible results when I enter the radius in m²/cm²
My calculation:
Volume: R = (3 * V / 4 * pi) ^ 1/3;
V = 1cm³ = 0.000001m³
R= 0,0133067m³
Or
R = (A / pi²) ½; A = 1cm² = 0.0001m²
R=0,0056m²
0n the pictures you can see that the results (1cm³) are not correct?! With R=0,0133067m³ the output is 46 Points…
Where is the mistake in my mind?
- Attachments
-
- cc_1cm.png (38.61 KiB) Viewed 6414 times
-
- CC_frage.png (79.91 KiB) Viewed 6416 times
Re: surface density
The units should be the same as the one in which the point coordinates are expressed in.
Looking at your snapshots, it looks like it's in meters?
Looking at your snapshots, it looks like it's in meters?
Daniel, CloudCompare admin
Re: surface density
yes the point coordinates and the Snapshots are in meters.
Re: surface density
For the 'population' density, you input the radius only (not the surface). So what do you have for R = 0.01?
Daniel, CloudCompare admin
Re: surface density
On the picture is my result.
The input with 0.01 m is not satisfactory. At least if I count the points manually.
The input with 0.01 m is not satisfactory. At least if I count the points manually.
- Attachments
-
- CC_Ausschnitt1.png (140.42 KiB) Viewed 6390 times