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# intersection point of three spheres

## intersection point of three spheres

1 How do I make a sphere at a given x,y,z and a given radius ?

2 Given three such random spheres,and if they intersected at only one point,
can the calculator find the intersection point not just visually, but actually manifest the x,y,z point to say 5 decimals?

Thanks for any help you can offer.

steve waterman

Posts : 13
Join date : 2012-03-21

## Re: intersection point of three spheres

1) The general formula is (x-pX)^2+(y-pZ)^2+(z-pZ)^2=1 where center of sphere would be at (pX, pY, pZ). But our GC3 software doesn't support implicit formulas yet (it will in the next release) so you need to rearrange that formula to z=(1-(x-pX)^2-(y-pZ)^2)^0.5+pZ

2) Currently GC3 does not compute intersection but we hope to add this feature to next release.

_________________
Saeid Nourian, Ph.D.
Founder of Runiter Company
http://www.runiter.com

runiter
Admin

Posts : 135
Join date : 2007-12-05
Age : 39
Location : Boston, MA, USA

## great, that worked. How does one make additional spheres?

Admin wrote:1) The general formula is (x-pX)^2+(y-pZ)^2+(z-pZ)^2=1 where center of sphere would be at (pX, pY, pZ). But our GC3 software doesn't support implicit formulas yet (it will in the next release) so you need to rearrange that formula to z=(1-(x-pX)^2-(y-pZ)^2)^0.5+pZ

2) Currently GC3 does not compute intersection but we hope to add this feature to next release.

Thanks for the rapid reply.
z=(1-(x-pX)^2-(y-pZ)^2)^0.5+pZ
and its partner
z=-(1-(x-pX)^2-(y-pZ)^2)^0.5+pZ

it appears that the 1 is actually the radius, excellent. That allowed me to place a sphere anywhere and to control the radius.

I tried to add a second sphere and could only get the first 1/2 of it. I would get an indexedquadarray illegalindexcount when I enter the second portion. I used x = and x =, and also tried y = and y =. I tried another z= and z=, but that did not succeed.

How would one code for a second and third sphere, please?

steve waterman

Posts : 13
Join date : 2012-03-21

## Re: intersection point of three spheres

[quote="steve waterman"]
Admin wrote:1) The general formula is (x-pX)^2+(y-pZ)^2+(z-pZ)^2=1 where center of sphere would be at (pX, pY, pZ). But our GC3 software doesn't support implicit formulas yet (it will in the next release) so you need to rearrange that formula to z=(1-(x-pX)^2-(y-pZ)^2)^0.5+pZ

2) Currently GC3 does not compute intersection but we hope to add this feature to next release.

Thanks for the rapid reply.
z=(1-(x-pX)^2-(y-pZ)^2)^0.5+pZ
and its partner
z=-(1-(x-pX)^2-(y-pZ)^2)^0.5+pZ

it appears that the 1 is actually the radius, excellent. That allowed me to place a sphere anywhere and to control the radius.

I tried to add a second sphere and could only get the first 1/2 of it. I would get an indexedquadarray illegalindexcount when I enter the second portion. I used x = and x =, and also tried y = and y =. I tried another z= and z=, but that did not succeed.

How would one code for a second and third sphere, please?

added -

aha, I got it. It was indeed another z= and z=. So I an fine and you need not respond to this question, as I did make a couple of spheres.
Any rough idea when a new version that could detect the intersection point might be possible?

steve waterman

Posts : 13
Join date : 2012-03-21

## Re: intersection point of three spheres

[quote="steve waterman"]
steve waterman wrote:
Admin wrote:1) The general formula is (x-pX)^2+(y-pZ)^2+(z-pZ)^2=1 where center of sphere would be at (pX, pY, pZ). But our GC3 software doesn't support implicit formulas yet (it will in the next release) so you need to rearrange that formula to z=(1-(x-pX)^2-(y-pZ)^2)^0.5+pZ

2) Currently GC3 does not compute intersection but we hope to add this feature to next release.

Thanks for the rapid reply.
z=(1-(x-pX)^2-(y-pZ)^2)^0.5+pZ
and its partner
z=-(1-(x-pX)^2-(y-pZ)^2)^0.5+pZ

it appears that the 1 is actually the radius, excellent. That allowed me to place a sphere anywhere and to control the radius.

I tried to add a second sphere and could only get the first 1/2 of it. I would get an indexedquadarray illegalindexcount when I enter the second portion. I used x = and x =, and also tried y = and y =. I tried another z= and z=, but that did not succeed.

How would one code for a second and third sphere, please?

added -

aha, I got it. It was indeed another z= and z=. So I an fine and you need not respond to this question, as I did make a couple of spheres.
Any rough idea when a new version that could detect the intersection point might be possible?

added - I have a very tricky question. Given two x,y,z points, I want to draw a hyperboloid between them.
That is, there is a numeric value of say, 3.6 that represents the constant distance that some point is closer to point A than that point is to point B. Any thoughts on this?
Noting that if this value was 0, then this would result in the plane 1/2 way between the point A and point B.

steve waterman

Posts : 13
Join date : 2012-03-21

## Re: intersection point of three spheres

steve waterman wrote:Any rough idea when a new version that could detect the intersection point might be possible?
We don't have a release date in mind yet but it could take 6 months.

_________________
Saeid Nourian, Ph.D.
Founder of Runiter Company
http://www.runiter.com

runiter
Admin

Posts : 135
Join date : 2007-12-05
Age : 39
Location : Boston, MA, USA

## Re: intersection point of three spheres

Admin wrote:
steve waterman wrote:Any rough idea when a new version that could detect the intersection point might be possible?
We don't have a release date in mind yet but it could take 6 months.

Thanks. I added onto the previous post, and likely you posted without seeing the addition about drawing hyperboloids.

steve waterman

Posts : 13
Join date : 2012-03-21

## Re: intersection point of three spheres

I now also see that in order to do parametric equations one needs more than just the free version. So, I have no remaining questions.

steve waterman

Posts : 13
Join date : 2012-03-21

## Re: intersection point of three spheres

Just a final thought.
I think your spheres would be better shown as complete solids with a proper opacity [ user controlled ].
Indeed, one cannot visualize intersection points where the sphere surfaces are vacant.
Perhaps opacity for each sphere is available now?, is there a way to make the entire sphere now?

steve waterman

Posts : 13
Join date : 2012-03-21

## Re: intersection point of three spheres

steve waterman wrote:I have a very tricky question. Given two x,y,z points, I want to draw a hyperboloid between them.
That is, there is a numeric value of say, 3.6 that represents the constant distance that some point is closer to point A than that point is to point B. Any thoughts on this?
I'm not sure how to do this. If you have a working formula in mind I can tell you how to translate it to work in GC3 but other than that I can't be of much help in this case.

runiter
Admin

Posts : 135
Join date : 2007-12-05
Age : 39
Location : Boston, MA, USA

## Re: intersection point of three spheres

steve waterman wrote:Just a final thought.
I think your spheres would be better shown as complete solids with a proper opacity [ user controlled ].
Indeed, one cannot visualize intersection points where the sphere surfaces are vacant.
Perhaps opacity for each sphere is available now?, is there a way to make the entire sphere now?
Yes we're aware of that limitation with spheres and when we allow implicit equations we also hope to render spheres as a single continuous mesh. Should be available in the next version as well.

runiter
Admin

Posts : 135
Join date : 2007-12-05
Age : 39
Location : Boston, MA, USA

## Re: intersection point of three spheres

Admin wrote:
steve waterman wrote:I have a very tricky question. Given two x,y,z points, I want to draw a hyperboloid between them.
That is, there is a numeric value of say, 3.6 that represents the constant distance that some point is closer to point A than that point is to point B. Any thoughts on this?
I'm not sure how to do this. If you have a working formula in mind I can tell you how to translate it to work in GC3 but other than that I can't be of much help in this case.

wherever A[x^2 + y^2 +z^2 ]^0.5 + some given lineal distance = B[x^2 + y^2 +z^2 ]^0.5 should do the trick, I think.
This sounds like a job that would be handled by the equation function of the new release.

Again, thanks so much for all the rapidly returned feedback.

steve waterman

Posts : 13
Join date : 2012-03-21

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