Minimum distance between a disk in 3d space and a point above the disk

In summary, the conversation discusses finding the minimum distance between the perimeter of a disk in 3D space and a point above the disk. Various methods are suggested, including using vector addition and projecting the point onto the x,y plane. However, it is ultimately determined that the distance between a point and a circle can be found by subtracting the radius of the circle from the distance between the center of the circle and the point. This method can be extended to 3D space by considering the radius of the circle as the distance between the center and the point, rather than the radius of the circle itself.
  • #1
CraigH
222
1
Hi all,
How can I calculate the minimum distance between the perimeter of a disk in 3d space and a point above the disk? (the point can be inside or outside the area above the disk)

I've been trying to work this out for a while, but I'm getting no where.

For example, a point at (1,1,1) and a disk with center (0,0,0) and radius 0.5. The distance between the centers is:

[tex]\sqrt{(1-0)^{2}+(1-0)^{2}+(1-0)^{2}} = \sqrt{3} \approx 1.73[/tex]

But how can I work out the shortest distance from the point to a point on the perimeter of the disk?

Thanks!

http://imageshack.us/a/img838/5558/aujg.jpg [Broken]
 
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  • #2
You could try vector addition:

Rcenter + Rmin + Rradius = 0 and so Rmin = - (Rcenter + Rradius)
 
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  • #3
Thanks! That seems obvious now you've mentioned it aha. I was trying to do it using trigonometry.
 
  • #4
Although shouldn't it be...

Rmin = Rradius - Rcenter

so in the example i gave

Rmin = (0.5,0,0) - (1,1,1)

Rmin = (-0.5,-1,-1)

so it has length 1.5
 
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  • #5
Also, I've just realized this only works when Rradius lies on the x axis. How would you find the vector Rradius (the vector between the disk center and min point on the perimeter) if it does not lie on the x axis?
 
  • #6
I think I've figured this out, it would help if someone could verify if I'm doing this correctly though.

There is a particle at [5,5,5] above a disk with center [0,0,0] and radius 3. The disk is not tilted: It lies on the x,y plane.

To calculate the vector between the disk center, and the point on the perimeter of the disk that is closest to the particle:

1) calculate unit vector from center to particle

unitvec=[5,5,5]/|[5,5,5] = [0.5774, 0.5774, 0.5774]

2) set the z element to 0, and multiply by the radius

vec1=3*[0.5774, 0.5774, 0] = [1.7321,1.7321,0]

To calculate the vector between the particle and the point on the perimeter of the disk that is closest to the particle:

1) subtract the vectors

[5,5,5]-[1.7321,1.7321,0] = [3.2679, 3.2679, 5.0000]

The magnitude of this is the minimum distance. This is 6.8087Am I making any mistakes here?
 
  • #7
Your step 2 is erroneous. Your vector [1.7321,1.7321,0] does not represent a point on the perimeter of the disk.
 
  • #8
Yeah I think the problem here is step two where you set the z coordinate to zero and assume that unitvec is still a unit vector. [0.5774, 0.5774, 0.5774] is a unit vector in 3d space, but [0.5774, 0.5774, 0] is not a unit vector on the x,y plane on which the disk lies. You can see for yourself if you evaluate (0.5774)^2+(0.5774)^2.
 
  • #9
As I understand the problem, you need to find the distance between some point in 3D space and the perimeter of a disk (i.e., a circle). You do not have to find the point on circle that is closest to the point in question to solve this problem.

Break the problem into parts. Start with the 2D version of the problem, finding the distance between a point [x,y] and a circle of radius r centered at the origin. Hint: You do not need to solve for the coordinates of the point on the circle that is closest to the point [x,y] to determine the distance between this point and the circle.
 
  • #10
hddd123456789 said:
the problem here is step two where you set the z coordinate to zero and assume that unitvec is still a unit vector.

You're right. I think this will be fixed if I project the particle position vector (the vector pointing from the disk center to the particle) onto the x,y plane by setting the z element to 0, and then normalising it, instead of normalising it and then projecting it onto the x,y plane.

so this will give a vector that points towards the position on the x,y plane that the particle is above, and it will have a magnitude of 1. So multiplying this by the radius will give the vector that points to the min point on the perimeter of the disk.
 
  • #11
You do not need to solve for that point. The problem is much easier to solve by not finding the location of that point.
 
  • #12
I need to find the closest distance to the particle. This is so that I can later work the solid angle subtended by the disk. Once I have worked out this distance I can use a similar method to work out the maximum distance, and then using these I can work out the solid angle subtended by the disk.

D H said:
Hint: You do not need to solve for the coordinates of the point on the circle that is closest to the point [x,y] to determine the distance between this point and the circle.
I cannot think of a way of finding the distance between a point and a circle without first finding the coordinates of the closest point on the circle.

Unless I work out a general equation for the distance between coordinates on a circle and a point, and then find where the derivative of this equation equals zero to find the minimum.

However this method seems more difficult.
 
  • #13
Think of the point [x,y] as being one point on a circle of some radius R with origin at the center.

What is the distance between two concentric circles, one of radius R and the other of radius r?
 
  • #14
D H said:
Think of the point [x,y] as being one point on a circle of some radius R with origin at the center.

What is the distance between two concentric circles, one of radius R and the other of radius r?

That distance will be R - r.

so if [x,y] is a point on the circle with radius R, and [a,b] is the location of the particle, then r will equal (a^2+b^2)^0.5

so the minimum distance between the circle and particle will be R-r

I can't work out how to use this method so that it will work in 3 dimensions though
 
  • #15
CraigH said:
That distance will be R - r.
Almost correct. What if R<r?
I can't work out how to use this method so that it will work in 3 dimensions though
Look at your own diagram. You have a triangle in it. One vertex is on the circle, another your point [x,y,z], the third is on the plane. One side is labeled Rmin, the side opposite the point [x,y,z] is a solid yellow-green line, and the third side is a dashed yellow-green line. The length of this third side is easy. You just figured how how compute the length of the side marked with a solid yellow-green line. All that is left is to compute the length of the side labeled Rmin.

What kind of triangle is this? (In other words, what is the angle between those two yellow-green sides?)
 
  • #16
D H said:
Almost correct. What if R<r?

Oh yeah, ((R-r)^2)^0.5 then.

D H said:
Look at your own diagram. You have a triangle in it. One vertex is on the circle, another your point [x,y,z], the third is on the plane. One side is labeled Rmin, the side opposite the point [x,y,z] is a solid yellow-green line, and the third side is a dashed yellow-green line. The length of this third side is easy. You just figured how how compute the length of the side marked with a solid yellow-green line. All that is left is to compute the length of the side labeled Rmin.

What kind of triangle is this? (In other words, what is the angle between those two yellow-green sides?)

Ah okay, so I can just use pythagoras. Let me try my example again to see if I've got it.

There is a particle at [5,5,5] above a disk with center [0,0,0] and radius 3

R = 3

r = ( (5^2)+(5^2) +(0^2) )^0.5 = 7.071

distance between concentric circles = ((3-7.071)^2)^0.5 = 4.071

height (the dashed line) = 5

so Rmin = ( (4.071^2) + (5^2) )^0.5 = 6.448

Is this correct? I hope it is because this method seems much easier, thanks!
 
  • #17
That is correct.
 
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1. What is the minimum distance between a disk in 3D space and a point above the disk?

The minimum distance between a disk in 3D space and a point above the disk is equal to the shortest distance between the point and the disk's surface. This can be calculated using the disk's radius and the coordinates of the point.

2. How do you calculate the minimum distance between a disk in 3D space and a point above the disk?

To calculate the minimum distance, you can use the formula d = |h - r|, where d is the minimum distance, h is the height of the point above the disk's center, and r is the radius of the disk.

3. Can the minimum distance between a disk in 3D space and a point above the disk be negative?

No, the minimum distance cannot be negative as it represents the shortest distance between the point and the disk's surface. If the point is directly above the disk's center, the minimum distance would be equal to the disk's radius.

4. How does the size of the disk affect the minimum distance between the disk and a point above it?

The size of the disk, represented by its radius, directly affects the minimum distance between the disk and a point above it. A larger disk will have a larger radius, resulting in a larger minimum distance.

5. Can the minimum distance between a disk and a point above it be greater than the disk's radius?

Yes, the minimum distance can be greater than the disk's radius if the point is located far enough above the disk's center. In this case, the minimum distance will be equal to the height of the point above the disk's center minus the disk's radius.

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