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Patriot_1776
Patriot_177622-Sapphire II
22-Sapphire II
March 6, 2012
Solved

For fun: How many points does it take to define a sphere's radius?

  • March 6, 2012
  • 6 replies
  • 3005 views

Pat 1: How many points in 3D space does it take to define the radius?

Part 2:

a. If given the points, how would you model it?

b. What is the resulting radius?

c. What is the location of the radius from the default CS?

Have fun!


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Best answer by DavidHerman

This is easier. No math involved. For 4 points, A,B,C,D.

1. Create a plane through 3 points, ABC

2. Sketch a circle through the three points A,B,C

3. Create an axis through the center of the circle

4. Create a plane through the axis and point D

5. Sketch a circle through point D and the intersection points of the circle and plane.

This sketch is the circumference of the sphere.

For the points you gave, the center is 32.93, 7.46, 3.02 and the radius is 32.39

6 replies

Patriot_1776
Patriot_177622-Sapphire IIAuthor
22-Sapphire II
March 7, 2012

Nobody wants to take a stab at this?

C
Constantin15-Moonstone
15-Moonstone
March 7, 2012

ok, not sure if this is a trick question...

Part 1: (if we are talking of a "radius" as in circle segment) 3

Part 2:

a. plane through three points, sketcher, three point arc

b. depends on the given points

c. see b.

D
DavidHerman10-Marble
10-Marble
March 7, 2012

Part 1: 4

(x - xo)^2 + (y - yo)^2 + (z - zo^2) = R^2

xo, yo, zo, and R are unknowns, so 4 points will give you 4 equations to solve for the 4 unknowns.

Part 2:

a. Center your sketch at xo, yo, zo; draw an arc of radius R; revolve.

b. R

c. Distance of center of sphere to default CS is sqrt(xo^2 + yo^2 + zo^2)

Patriot_1776
Patriot_177622-Sapphire IIAuthor
22-Sapphire II
March 8, 2012

4 points it is. Good job! I can't vouch for the math, but I'll post my model shortly.

Before that, I'll list the points based off the default CS:

!

! DATUM POINT ARRAY DATA FILE

!

! Comment lines should begin with an exclamation mark (!).

! You may have as many comment lines as you wish.

! There must be at least one space between values.

!

! Enter values with respect to datum arrays' coordinate system:

!

!CARTESIAN coordinates:

! X Y Z

!

1.000 2.000 3.000

2.000 4.000 12.000

3.000 16.000 12.000

60.000 25.000 0.000

So, relative to the default CS, what's the radius, and where is it located in X, Y, and Z?

D
DavidHerman10-MarbleAnswer
10-Marble
March 12, 2012

This is easier. No math involved. For 4 points, A,B,C,D.

1. Create a plane through 3 points, ABC

2. Sketch a circle through the three points A,B,C

3. Create an axis through the center of the circle

4. Create a plane through the axis and point D

5. Sketch a circle through point D and the intersection points of the circle and plane.

This sketch is the circumference of the sphere.

For the points you gave, the center is 32.93, 7.46, 3.02 and the radius is 32.39

Patriot_1776
Patriot_177622-Sapphire IIAuthor
22-Sapphire II
March 12, 2012

Good job. I actually did it a little differently, but very close.

Funny, I realized I'd posted this before as "noname" when I first did it in '07. I recently ran across that thread again. Only you and Grahame did it.

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