## Find circle radius & circle center in 3D from (3) XYZ pts

## Find circle radius & circle center in 3D from (3) XYZ pts

(OP)

GIVEN: X-Y-Z coordinates of 3 points in 3D space. FIND: Radius of circle on those 3 points & Circle center co-ordinatesBookmark:

Question: Can someone provide a mathematical step by step procedure to calculate and determine a circle radius and the center point coordinates from three GIVEN 3D points with X-Y-Z coordinates?

- I have searched the web extensively, and found lots of proposed solutions, but my college math is quite rusty and after several days, I still can not come up with the correct detailed solution. I can do this easily from three points in 2D, in Excel using three simultaneous equations with TWO variables (i.e. find circle center @ h, j). But I can't get it to work in 3D with a similar spreadsheet approach with THREE variables (i.e. find circle center @ h-j-k). Seems I keep going in mathematical circles (no pun intended) with a modified Excel approach with the three quadratic equations.

- Is it possible to solve this mathematically with (3) simultaneous quadratic equations? Or do I need a 4th equation, determinates, etc?. Seems when I try the double elimination (or substitution) of variables approach, I keep coming up with unsolvable, or meaningless equations. With as much time as we have spent on this problem, we would be happy to compensate someone for their time to solve this for us.

- We have several drawings with bent pipes in 3D where we have only THREE 3D points given to us. We need to be able to determine the TRUE bend radius of the pipes. These are special alloy pipes @ $25k each, so we want to be sure we have the correct 2D bend radius before we bend them. I'd prefer an approach that could be put into a spreadsheet to make reiterative calcs go quickly. I'd prefer to have a detailed step-by-step solution vs. being directed to one of the web solutions. I have seen most of them, and still am stuck on this.

Thanks, MWP

Question: Can someone provide a mathematical step by step procedure to calculate and determine a circle radius and the center point coordinates from three GIVEN 3D points with X-Y-Z coordinates?

- I have searched the web extensively, and found lots of proposed solutions, but my college math is quite rusty and after several days, I still can not come up with the correct detailed solution. I can do this easily from three points in 2D, in Excel using three simultaneous equations with TWO variables (i.e. find circle center @ h, j). But I can't get it to work in 3D with a similar spreadsheet approach with THREE variables (i.e. find circle center @ h-j-k). Seems I keep going in mathematical circles (no pun intended) with a modified Excel approach with the three quadratic equations.

- Is it possible to solve this mathematically with (3) simultaneous quadratic equations? Or do I need a 4th equation, determinates, etc?. Seems when I try the double elimination (or substitution) of variables approach, I keep coming up with unsolvable, or meaningless equations. With as much time as we have spent on this problem, we would be happy to compensate someone for their time to solve this for us.

- We have several drawings with bent pipes in 3D where we have only THREE 3D points given to us. We need to be able to determine the TRUE bend radius of the pipes. These are special alloy pipes @ $25k each, so we want to be sure we have the correct 2D bend radius before we bend them. I'd prefer an approach that could be put into a spreadsheet to make reiterative calcs go quickly. I'd prefer to have a detailed step-by-step solution vs. being directed to one of the web solutions. I have seen most of them, and still am stuck on this.

Thanks, MWP

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

First use the 3d points to define the plane, then on that plane use the same 3 points to define the circle.

Note that you haven't beaten the requirement that it takes 4 triads to define a sphere, as this merely finds a section through an infinite number of spheres.

Cheers

Greg Locock

SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

Sounds an interessting and challenging problem. Are the three points given always the same three points on the arc i.e are they always 45deg apart or some other known constant? This may be the way to simplify the solution into a planar (2D) problem.

Mark Hutton

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

I presume that all bends that you want to determine are bends in one plane, and that the points given are points on the bend radius in that plane?

If you then first determine the plane and find the xy coordinates for the three points in that plane you should be able to go back to your 2D solution.

Are all the bends 90 degrees?

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

ANALYTICALLY:

A circle is a planar geometrical object, so you must resolve your 3D curve to 2D.

From my CRC Standard Math Tables 26th Edition (Analytic Geometry)

3-point form {x1,y1}, {x2,y2}, {x3,y3}:

solve the simultaneous equations matrix:

| (x^2 + y^2) x y 1 |

| (x1^2 + y1^2) x1 y1 1 | = 0

| (x2^2 + y2^2) x2 y2 1 |

| (x3^2 + y3^2) x3 y3 1 |

If you're still stumped, go to the library and check out a CRC math book...it's all in there.

TygerDawg

Blue Technik LLC

Virtuoso Robotics Engineering

www.bluetechnik.com

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

The OP said that these are bends defined by 3 points.

As such, they are coplaner.

Establish a Plane using said points, then find the radius using same.

As for the Excel-fired solution to h-j-k, I'm not sure what works....

Here's something I found on the 'net

h

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

of course, you could put the three co-ords into Auto-Cad and it'll solve this for you ...

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

If you have "several" and each is ~ $25k, your time spent on trying to get Excel to work might be better spent just modeling the parts up, and obtaining the h-j-k co-ordinate if this is indeed a requirement at all.

I've done some tubing modeling in both AutoCad & SW where the 3D points are given for every transition from straight to bend, and back to straight; each bend having a unique value not explicitly given, but geometrically correct.

(All bends being tangent to the straights)

None of this is in reference to a sphere.

It goes rather quickly, once you get onto it.

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

For an analytic solution, four equations are needed. Expanding on the previous posts, there are the three obvious equations plus a fourth that constrains all points to a plane. I skipped the tensor transforms and tried setting up something in 4eqns/4unkn. Here's a draft but be forewarned, its ugly. Maybe you can use it (or lose and go the CAD route).

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

In 3D:

The center is on plane ABC.

The center is on the bisector planes of segments AB, BC, AC (you would only need two of these planes).

The center of the circle is at the intersection of three planes: ABC, bisector of AB, and bisector of BC.

Honesty may be the best policy, but insanity is a better defense.http://www.EsoxRepublic.com-SolidWorks API VB programming help

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

There are 3 unknowns, so only 3 equations should necessary. However, pathological configurations of the 3 points amy result in a poor solution.

Have you tried to structure the three 3D distance equations from the center, and have Excel solve for when all 3 distances are equal? You don't, at this point, really care what the equation of the sphere looks like, you simply need to find the center.

However, as with GPS, you could get a poor geometrical dilution of precision, particularly if the 3 points are clustered within a small portion of the sphere.

TTFN

FAQ731-376: Eng-Tips.com Forum Policies

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

The center is found by first finding the plane that is the bisector of any two points,say point 1 and point 2. Then find the plane of the bisector of point 2 and point 3. Now since the intersection of these planes is a LINE and EVERY point on that line is equidistant to the 3 points so that, there is no unique center and therefore no unique sphere. Now you need a 4th point in space to determine that unique sphere. You get it obviously by finding a 3rd plane that bisects the line from that 4th point to any of the other points; the intersection of the 3 planes so formed is the center. which solves your problem. You get it by writing 4 equations:

(X-Xj)^2 +(Y-Yj)^2+(Z-Zj)^2=R^2, for j=1,2,3,4

where

X,Y,Z are the unknown coordinates of the sphere center

and R is the unknown radius.

4 equations and 4 unknowns should get the solution unless the 4 points are co-planer.

r

where

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

1) I know we need 4 points to define a sphere. But we are not looking for anything spherical. We are only looking for a radius and the ceenter of that radius, as determined by the three XYZ defined points (non-colinear).

2) The 3 pts do define a triangle on a plane. The three points of any triangle will lie on a unique circle.

3)My primary focus was to see if anyone had a mathematical (vs. geometrical) solution that could be put into a spreadsheet. As I stated in my 1st post ... my college math is quite rusty. That's probably why I was having trouble coming up with the math way to define the planes, bisectors, and then solve the problem. I think I am close with the simultaneous equations, but as someone stated above, I think I need one more point to make the 4th equation.

4) More clearly, the points we are given are totally random points in 3D space from some a common set of XYZ axes with 0,0,0 origin. The three points for each pipe are given:

Pc (starting point of curvature, tangent to the bend centerline, and the next straight section),

Pt (Ending point of the curved section, tangent to the bend centerline, and to the next straight section), and

Pi (point of intersection of the two tangents above)

The curved section is a constant radius (once transformed into a 2D plane). Out of plane view, it is a slight ellipse, or a straight line (if viewed parallel to the planar view.

- After bending, the pipe is tilted, offset, and rotated into positon to put the 3 points into proper position. Then, these are welded to large fabrications, that are shipped 3000 miles away. And the end points have to mate up excactly to someone elses pipe work, using the same drawings.

- The arc lengths and arc angles vary from pipe to pipe.

- I am sure that this be done graphically in Solid Works, but them I have no way to mathematicallly verify my CAD guys work is dimensinally correct.

- Lots of comments about this being a 2D planar solution ... and it is, but I don't know how to convert the 3D points into a 2D plane. THis is kind of what I commented on in my first post ... rusty math that is.

- How is the plane mathematically "defined" by the three

points? I can probablly find this ok.

- I think I have my old college CRC book ... good suggestion .. I will look at that also.

Willedawg ... I had already found that mathforum site, and had studied that particular solution .. but was still a little fuzzy on that onee. Seems more complex than it needs to be ... but we are not trying to define a sphere.

I will pass your info to my SWks guy ... we don't have ACAD.

rb1957 ... that Engr.com link apppears to be for integer solutions, and only in 2D. I don't see the relevance. However, the Engr.com link from Bestwrench appears to give me the 4th equation I was missing to plug these into a simple simultaneous equation solution in Excel. I will play with that one too. I did a ton of google searching and never hit on the Engineering.com site (not sure why).

Zekeman ... again, I am looking for the circle, not the sphere, and I only have 3 points, so I don't think this will work for me.

Thanks again to everyone. Big help overall. Any other comments still welcome.

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

X Y Z

1 17 300 4

2 -99 56 2

3 8 5 -33

Centre 20.6 148.6 14.4

Radius 151.8

Cheers

Greg Locock

SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

But, you must know that in order to get a circle that is tangent to lines formed by pts 1 and 2 at pt 1 and pts 2 and 3 at pt 3, the lines 1-2 and 2-3 must be of equal length.

First write the equation of the plane (pts1,2,3)

eq1 x+ay+ bz=c

where a,b and c are determined by entering the coordinates of the 3 points, giving 3 equations whose solution of a,b,c determines the plane of eq1.

Now we get the direction cosines of the 2 lines 1-2 and 2-3

which are proportional to

line 1-2

(x1-x2),(y1-y2),z1-z2)

line 2-3

(x2-x3),y2-y3),z2-z3)

Next, we draw lines from a point X,Y,Z on the plane of eq1 to pt 1 and pt 3. Their respective direction cosines are proportional to:

(X-x1),(Y-y1), Z-z1) line Q X,Y,Z to x1,y1,z1

(X-x3),Y-y3),Z-z3) line R X,Y,Z to x3,y3,z3

Since line Q is perpendicular the line 1-2 at point 1 and line R is perpendicular to line 2-3 at point 3,(definition of tangency), the dot product of these respective line pairs are zero

Expanding

(X-x1)*(x1-x2)+(Y-y1)*(y1-y2)+(Z-z1)*(z1-z2)=0

(X-x3)*(x2-x3)+(Y-y3)*(y2-y3)+(Z-Z3)*(z2-z3)=0

From eq 1, X,Y, Z being in the plane, is constrained by

X+aY+bZ=c

Now we have 3 eq and 3 unknowns in X Y and Z

The solution X,Y,Z is the center of the circle.

A note of caution. I REPEAT, that you MUST have the 2 lines 1-2 and 2-3 of equal length; otherwise you wlll get wrong answers.

Hope this helps

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

now you know two tangents, (X1Y1-X2Y2) and (X1Y1-X3Y3), and two points of tangency, (X2Y2) and (X3Y3). determine the normals to the tangents thru the points of tangency. their intersection is the centre of the arc.

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

http://

-tg

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

you only need the two tangents and one point of tangency, as the centre is on the bisector of the two tangents.

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

Given: tangent points A & C, intersection point B...

•Center of arc lies on the bisector of ABC

•Bisector can be determined by line from B to midpoint of segment AC.

•Radius is distance from A or C to bisector.

•Center is found on line perpendicular from A or C to bisector (may also be found w/ point-to-curve distance calculation).

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

sqrt[X-x1)^2+(Y-y1)^2+(Z-z1)^2]

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

"i'd simply zeke's approach by redefining the three (xyz) points in terms of their common plane, (XY), removing one dimension.

now you know two tangents, (X1Y1-X2Y2) and (X1Y1-X3Y3), and two points of tangency, (X2Y2) and (X3Y3). determine the normals to the tangents thru the points of tangency. their intersection is the centre of the arc."

-------------------------------------------------------------

Do you really think you simplify the solution by working in the x-y plane?

What about the transformation into the x-y plane? I don't really think so.

The 3D solution is fairly simple these days thanks to the quick ability of any of the math programs to solve linear equations.

I could easily set up the solution in EXCEL - first solving my 2x2 set and then the final 3x3 set.

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

Cheers

Greg Locock

SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

zeke,

actually, yes, i do think working in 2D is easier than 3D. i think working with co-ords like (0,0), (x,0), and (x1,y1) is easier. i think looking on the arc for the normal (the true view) is easier to visualise. admittedly, the math is moved to the transformation matrices, but none of this math is "hard".

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

Greg Locock

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

R and included angle I. To find position of center you will need to solve three simultaneous equations.

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

- Sorry for the delay to say thanks to all for the great responses to this thread. I have not completed this yet, but am very confident that we have sufficient information to complete the solution I was after. So thanks again to all. Very nicely done! MWP

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

Finding the centre and radius of the circumscribed circle of a triangle (ABC) is something I first did in the mid-1980s — in BASIC, I'm retired now but I've still got the notes written at the time. The normal form of the plane equation wasn't used — just the direction cosines of the axis system that the three points are in. The diameter of the circle that fits the corners of the triangle is the length of

anyside divided by the sine of its opposite angle; the centre of the circle is found by solving one more triangle.The following code is modified BASIC; if it's pasted into an Excel VBA macro it will run — it is basic stuff in more ways than one. I'm sure there are slicker ways of doing this task, but I've used direction cosines; the code has been run with various values (and signs) for the three points — and has outputted the correct (so far...) radius and radius centre coordinate each time. (I've had it checked against a CATIA model)

I've posted the code because an original requirement was for a solution that could be pasted in and run.

## CODE

Range("A1:Z512").NumberFormat = "###0.000000"

' ———————————————————————————————————————————————————————————————————

' Given three points in space (A,B,C)

Xa = 47: Ya = -61: Za = 137

Xb = -6: yb = 19: Zb = -28

Xc = 83: Yc = 40: Zc = -55

' Lengths of AB, AC, AC

AB = (((Xa - Xb) ^ 2) + ((Ya - yb) ^ 2) + ((Za - Zb) ^ 2)) ^ 0.5

BC = (((Xb - Xc) ^ 2) + ((yb - Yc) ^ 2) + ((Zb - Zc) ^ 2)) ^ 0.5

AC = (((Xa - Xc) ^ 2) + ((Ya - Yc) ^ 2) + ((Za - Zc) ^ 2)) ^ 0.5

' Direction cosines of AB(ABi,ABj,ABk)

ABi = (Xb - Xa) / AB

ABj = (yb - Ya) / AB

ABk = (Zb - Za) / AB

' Direction cosines of AC(ACi,ACj,ACk)

ACi = (Xc - Xa) / AC

ACj = (Yc - Ya) / AC

ACk = (Zc - Za) / AC

' Cosine of angle BAC

cosBAC = (AB ^ 2 + AC ^ 2 - BC ^ 2) / (2 * AB * AC)

AD = cosBAC * AC

CD = (AC ^ 2 - AD ^ 2) ^ 0.5

' Position of point D, which is C projected normally onto AB

Xd = Xa + (AD * ABi)

Yd = Ya + (AD * ABj)

Zd = Za + (AD * ABk)

' Direction cosines of CD(Cdi,CDj,CDk)

CDi = (Xc - Xd) / CD

CDj = (Yc - Yd) / CD

CDk = (Zc - Zd) / CD

' Direction cosines of normal to AB and CD

' — to be used for rotations of circle centre

Ni = (ABk * CDj) - (ABj * CDk)

Nj = (ABi * CDk) - (ABk * CDi)

Nk = (ABj * CDi) - (ABi * CDj)

' # Diameter of circumscribed circle of a triangle is equal to the

' the length of any side divided by sine of the opposite angle.

' This is done in a coordinate system where X is colinear with AB, Y is // to CD,

' and Z is the normal (N) to X and Y, and the origin is point A

' R = D / 2

sinBAC = (1 - cosBAC ^ 2) ^ 0.5

R = (BC / sinBAC) / 2

' Centre of circumscribed circle is point E

X2e = AB / 2

Y2e = (R ^ 2 - X2e ^ 2) ^ 0.5

Z2e = 0

' Transform matrix

' Rotations Translations

' ——————————————————————————————————————————————

' ABi , ABj , ABk Xa

' CDi , CDj , CDk Ya

' Ni , Nj , Nk Za

' ——————————————————————————————————————————————

' Position of circle centre in absolute axis system

X_centre = Xa + (X2e * ABi) + (Y2e * CDi) + (Z2e * Ni)

Y_centre = Ya + (X2e * ABj) + (Y2e * CDj) + (Z2e * Nj)

Z_centre = Za + (X2e * ABk) + (Y2e * CDk) + (Z2e * Nk)

' ———————————————————————————————————————————————————————————————————

Cells(1, 1) = "Radius =": Cells(1, 2) = R

Cells(3, 1) = "X_centre =": Cells(3, 2) = X_centre

Cells(4, 1) = "Y_centre =": Cells(4, 2) = Y_centre

Cells(5, 1) = "Z_centre =": Cells(5, 2) = Z_centre

End Sub

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

Shouldn't that formula be:

T/R = TAN(I/2) ... therefore, R = T / TAN(I/2)?

I plugged these formulas into Excel. When I ran it this way, it seems to come out correct. But let me know if you disagree.

Kapitan .. thanks for that post too ... on occasion, we do have the "3 points on the radius" problem. That will help. In this case, we only have the 2 points on the arc, and the point of intersection of the two tangents at those points.

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

I drew up a random 3-D Sketch in SW, obtained all the Points, Lengths, & curve data, etc....

then, to check these,

I used Chicopee's method in Excel (and substituting R = T / TAN(I/2) for formula #4 as DrMetal mentions...which worked for me too)

Fine so far. What stumps me is step #6. The formulas all look good and make sense, but how does one put them into the 3 simultaneous equations to solve for x4,y4,z4 ?

abs(R) = (x1-x4)^2 + (y1-y4)^2 + (z1-z4)^2

andabs(R) = (x2-x4)^2 + (y2-y4)^2 + (z2-z4)^2

abs(V) = abs(R)^2 + abs(T)^2

abs(V) = (x3-x4)^2 + (y3-y4)^2 + (z3-z4)^2

how is all this set up in Excel ?

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

the solution is (i think) very simple

given two points of tangency of an arc, and

given the point of intersection of these two tangents, then

two point form of a line (see wiki) gives you the equation of the tangents,

the two radii are normal to the tangents thru the points of tangency (normal to a line, slope/point form of a line, see wiki)

the center of the arc is the intersection of the two radii

the length of the radius is the distance from a point of tangency to the centre

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

- I have not finished going thru all this to solve for the circle center. But I believe the answer is here. I have an Excel SS set up to solve for Xc,Yc in 2D ... using 3 simulataneous equations, it works fine to solve for the 2 points (Xc,Yc) ... so I assumed I need 4 equations for the 3D points. But my math is a bit rusty too ... so I am still working on it ... just have not had the time yet.

The first 3 eq are easy:

Eq1: (X1-Xc)^2 + (Y1-Yc)^2 + (Z1-Zc)^2 = R^2

Eq1: (X2-Xc)^2 + (Y2-Yc)^2 + (Z2-Zc)^2 = R^2

Eq1: (X3-Xc)^2 + (Y3-Yc)^2 + (Z3-Zc)^2 = R^2

for ALL 3 points ON THE RADIUS

or

Eq2: (X2-Xc)^2 + (Y2-Yc)^2 + (Z2-Zc)^2 = V^2

for the Point of Intersection not on the radius

- We've already solved for R and V, so it looks easy, but when I try to solve for Xc, Yc, Zc, using only 3 equations, the math just runs me in circles and does not lead me to a solution.

- It's the lack of a 4th equation that is stumping me. I think we may have the answer in this thread, I just have not had time to do it. With 4 equations, you can solve for two variables 1st , then solve again for the one of these two variables in terms of the 2nd, and plug back in for the value of the 1st, and so on.

- What might also help would be to find the XYZ coordinates of the point of intersection of the circle arc and the bisector line between the line of POI (Point of Intersection of the 2 tangents) to the circle center. That would give a 3rd point on the arc too and might make it easier. But I have not looked at this yet.

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

i'll assume you didn't (couldn't?) follow my solution ...

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

- I thought that since we can solve for R and V, that the numbers could be used in the simultaneous equations to find the center point coordinates, but maybe not. Again, maybe my rusty math at work. I still haven't been able to put the solution together yet. Some related to time to work on this, but I don't see it clearly spelled out here.

- BTW ... is solving a set of simultaneus equations identical to solving similar problems using Cramers rule or similar?

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

http://

Including a spreadsheet with an on-spreadsheet solution, and two User Defined Functions that will find the 3D circle centre given either any 3 points on the circle, or two tangent points and the intersection of the tangents.

The method used for 3 points on the circle is:

1 Read the 3D coordinates for three points on the circle

2 Translate Point 2 and Point 3 for an origin at Point 1

3 Find polar coordinates of Point 3

4 Rotate Points 2 and 3 about the Z axis so that Point 3 is on the XZ plane

5 Rotate Points 2 and 3 about the Y axis so that Point 3 is on the X axis

6 Find angle of Point 2 from XY plane

7 Rotate Points 2 about the X axis so that Point 2 is on the XY plane

8 Find the XY coordinates of the mid-points of lines 1-2 and 3-2.

9 Find a second point on the perpendiculars through mid-points

10 Find the XY coordinates of the intersection of the perpendiculars. This is the centre of the circle.

11 Find the radius of the circle

12 Rotate and translate the centre point back to the original axes

13 Assign the 3D coordinates of the circle centre, and the circle radius, to the function return value as a 1×4 array

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

The functions give identical answers, and also take almost exactly the same time to run.

## CODE

' Based on "CircumscribedRadius_04() posted by "Kapitan" on the Eng-Tips Forum, 5 Jul 09

' http://eng-tips.com/viewthread.cfm?qid=248118&page=1

Dim XA As Double, YA As Double, ZA As Double

Dim XB As Double, YB As Double, ZB As Double

Dim XC As Double, YC As Double, ZC As Double

Dim AB As Double, BC As Double, AC As Double

Dim ABi As Double, ABj As Double, ABk As Double

Dim ACi As Double, ACj As Double, ACk As Double

Dim cosBAC As Double, AD As Double, CD As Double

Dim XD As Double, YD As Double, ZD As Double

Dim CDi As Double, CDj As Double, CDk As Double

Dim Ni As Double, Nj As Double, Nk As Double

Dim sinBAC As Double, R As Double, X2e As Double, Y2e As Double, Z2e As Double

Dim X_Centre As Double, Y_Centre As Double, Z_Centre As Double

Dim Cent(1 To 1, 1 To 4) As Double

If TypeName(Point1) = "Range" Then Point1 = Point1.Value2

If TypeName(Point2) = "Range" Then Point2 = Point2.Value2

If TypeName(Point3) = "Range" Then Point3 = Point3.Value2

' ———————————————————————————————————————————————————————————————————

' Given three points in space (Point1, Point2, Point3)

XA = Point1(1, 1)

YA = Point1(1, 2)

ZA = Point1(1, 3)

XB = Point2(1, 1)

YB = Point2(1, 2)

ZB = Point2(1, 3)

XC = Point3(1, 1)

YC = Point3(1, 2)

ZC = Point3(1, 3)

' Lengths of AB, AC, AC

AB = (((XA - XB) ^ 2) + ((YA - YB) ^ 2) + ((ZA - ZB) ^ 2)) ^ 0.5

BC = (((XB - XC) ^ 2) + ((YB - YC) ^ 2) + ((ZB - ZC) ^ 2)) ^ 0.5

AC = (((XA - XC) ^ 2) + ((YA - YC) ^ 2) + ((ZA - ZC) ^ 2)) ^ 0.5

' Direction cosines of AB(ABi,ABj,ABk)

ABi = (XB - XA) / AB

ABj = (YB - YA) / AB

ABk = (ZB - ZA) / AB

' Direction cosines of AC(ACi,ACj,ACk)

ACi = (XC - XA) / AC

ACj = (YC - YA) / AC

ACk = (ZC - ZA) / AC

' Cosine of angle BAC

cosBAC = (AB ^ 2 + AC ^ 2 - BC ^ 2) / (2 * AB * AC)

AD = cosBAC * AC

CD = (AC ^ 2 - AD ^ 2) ^ 0.5

' Position of point D, which is C projected normally onto AB

XD = XA + (AD * ABi)

YD = YA + (AD * ABj)

ZD = ZA + (AD * ABk)

' Direction cosines of CD(Cdi,CDj,CDk)

CDi = (XC - XD) / CD

CDj = (YC - YD) / CD

CDk = (ZC - ZD) / CD

' Direction cosines of normal to AB and CD

' — to be used for rotations of circle centre

Ni = (ABk * CDj) - (ABj * CDk)

Nj = (ABi * CDk) - (ABk * CDi)

Nk = (ABj * CDi) - (ABi * CDj)

' # Diameter of circumscribed circle of a triangle is equal to the

' the length of any side divided by sine of the opposite angle.

' This is done in a coordinate system where X is colinear with AB, Y is // to CD,

' and Z is the normal (N) to X and Y, and the origin is point A

' R = D / 2

sinBAC = (1 - cosBAC ^ 2) ^ 0.5

R = (BC / sinBAC) / 2

' Centre of circumscribed circle is point E

X2e = AB / 2

Y2e = (R ^ 2 - X2e ^ 2) ^ 0.5

Z2e = 0

' Transform matrix' Rotations Translations

' ——————————————————————————————————————————————

' ABi , ABj , ABk Xa

' CDi , CDj , CDk Ya

' Ni , Nj , Nk Za

' ——————————————————————————————————————————————

' Position of circle centre in absolute axis system

X_Centre = XA + (X2e * ABi) + (Y2e * CDi) + (Z2e * Ni)

Y_Centre = YA + (X2e * ABj) + (Y2e * CDj) + (Z2e * Nj)

Z_Centre = ZA + (X2e * ABk) + (Y2e * CDk) + (Z2e * Nk)

' ———————————————————————————————————————————————————————————————————

Cent(1, 1) = X_Centre

Cent(1, 2) = Y_Centre

Cent(1, 3) = Z_Centre

Cent(1, 4) = R

ArcCenP3_2 = Cent

End Function

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

maybe i'm being unusually thick, and i haven't gone through your solution line-by-line, however, if i have three points on a plane i can make to different arcs through them depending on the tangency condition. i'm surprised that a math procedure can determine the right choice ?

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

And yes, the spreadsheet has a function to find the circle defined by two tangent points and their intersection point (in fact it works with one tangent point and any point on the other tangent line past the tangent point). Thanks for reading :)

zekeman - as noted above my functions do address the two tangent question. If you read the posts you would also have seen that the calculation time using the approach of rotating the points onto a plane was almost identical to that using kapitan's routine, which used the direction cosine approach. To my mind the rotation approach makes it more intuitively obvious what is going on and why it works. I might have a go at putting your method into a function when I have some time, but not today.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

Just as he suggested, it was pasted into a Macro in VB and then Run. In order to use some other 3D points in space, I invoked the Macro, then used "Step Into" to edit in some different values; Ran it, noted the new values, then went back and compared to a SW drawing. All looked OK. Is this the "normal" way to do this?

Now,... how do I get the 2nd Code written by Doug in this thread to work ?

It starts out as a Function, not a Sub. How is this implemented or made into a Macro in Excel Basic ?

Or is some other Excel method/procedure needed?

I looked at the IP spreadsheet and downloaded it, but am having a hard time separating out only what's needed just for the circle and how to initialize it with the variable 3 point inputs.

Thanks for all the time and effort you have put into this.

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

It should work exactly like a built in function. The 'help' on user functions describes what to do.

Greg Locock

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

You can either follow the instructions below using the IP.xls spreadsheet, or if you prefer you can copy and paste the code into a Visual Basic module in a new spreadsheet, but if you do that you will need to also copy the functions Rotate() and IP(), which are called from the other functions. It's probably easier just to use the IP spreadsheet for a start.

To use any of the three functions enter the X, Y, and Z coordinates of three points on three separate lines. You can use any sheet. Suppose you enter the data in the ranges: A1:C1, A2:C2, and A3:C3, with row 1 and 3 being points on the circle and line 2 either an intermediate point on the circle or the tangent intersection point. Then to use the functions enter in any convenient cells:

=ArcCenT2IP(A1:C1, A3:C3, A2:C2)

=ArcCenP3(A1:C1, A2:C2, A3:C3)

=ArcCenP3_2(A1:C1, A2:C2, A3:C3)

Note the different order for ArcCenT2IP

In each cell the X coordinate of the circle centre will appear. ArcCenT2IP will give a different result to the other two of course.

The functions return an array; 1x4 for ArcCenP3 and ArcCenP3_2 and 2x4 for ArcCenT2IP. To get the other results select 4 adjacent columns and 1 or 2 rows, with the function entered in the top left cell, press F2 to enter edit mode, then press ctrl-shift-enter. There are examples of this on the IP spreadsheet.

An example of using the function called from another subroutine is given below:

## CODE

Dim TP1 As Variant, TP2 As Variant, TIP As Variant

Dim Cent As Variant, Res As Range

TP1 = Range("A1:C1").Value2

TP2 = Range("A2:C2").Value2

TIP = Range("A3:C3").Value2

Set Res = Range("A5:D6")

Cent = ArcCenT2IP(TP1, TP2, TIP)

Res = Cent

End Sub

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

- See what I got you into! Heck-of-a thread on this subject. I had no idea that it would be this "intriguing" to all the posters. Thanks for all of your input on this, and your emails too, and everyone else for the postings. We are getting ready to do the original PIPE developments and BENDS as described in earlier posts. FYI ... these are 8" x XXS (0.875"THK) 2205 DUPLEX SSTL PIPES x 20'L to bend and put into these 3D positions. We'll be using this thread to finally figure out and mathematically verify the correct 2D bends. Will post the final results of our success (or degree of success) later. Thanks again to all. Welcome to keep commenting if there is a need. Mike P.

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

Got the IP.xls to work.

Entered some values in a new Sheet and compared to SW drawing...accurate to 4 plcs.

"To get the other results select 4 adjacent columns and 1 or 2 rows, with the function entered in the top left cell, press F2 to enter edit mode, then press ctrl-shift-enter."

This part didnt seem to work for me. (maybe something I don't understand about the array function)

Instead of "select 4 adjacent columns", I selected the cell containing =ArcCenT2IP(A1:C1, A3:C3, A2:C2), then drag-selected 3 more columns to the right, but only 1 row, then did the F2 ctrl-shift-enter thing, and it filled the cells with the correct values for a 2pt circle with tangent lines intersecting at the 3rd pt.

For the next row down, I did the same thing, and it filled in the values for the 3pt circle situation

Doug, Kapitan and Greg, Thanks so much for the help on this topic and your patience with an Excel novice.

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

That's what I meant. I usually hold down the shift key and use the arrow keys to select. So I start on the cell with the function in it, hold down shift and press -> 3 times then F2 and ctrl-shift-enter. The mouse works just as well of course. I need to find a way to describe the process more clearly.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

- Quick question ... I must be doing something wrong ... when I plug in new numbers into the points for XYZ in your spreadsheet, the results all return the "#NAME" error, as if it can't read my inputs. Do I need to format the numbers that I enter a certain way? Or something else simple that I am doing wrong? I was just playing around with it a little tonight and could not get the Excel sheet to show numbers. As soon as I change even one of your sample entries, I get nothing but cells of errors. Sorry for the basic question, but I am still an Excel novice, and remember very little about basic and almost nothing about VB for that matter. Thanks again for the work. Mike P.

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

Sorry, I didn't read your post but I don't like rotating and unrotating stuff which to me is unnecessary.

My take is:

The center of the circumscribing circle is simply at the intersection of 3 planes, namely

1) the plane determined by the 3 pts 1,2,3 in space

2) the plane perpendicular to line 1-2 containing the midpoint of 1-2

3) the plane perpendicular to line 2-3 containing the midpoint of 2-3.

First write the equation of the plane (pts1,2,3)

eq1 x+ay+ bz=c

where a,b and c are determined by entering the coordinates of the 3 points, giving 3 equations whose solution of a,b,c determines the plane of eq1.(from my previous post)

Now we get the direction cosines of the 2 lines 1-2 and 2-3

which are proportional to

line 1-2

(x1-x2),(y1-y2),z1-z2)

line 2-3

(x2-x3),y2-y3),z2-z3)

Next we write the equation of the plane perpendicular to line 1-2

eq2 (x1-x2)X+(y1-y2)Y+(z1-z2)Z=D1

where D1 is determined from the midpoint location (x1+x2)/2,y1+y2)/2,(z1+z2)/2 in this plane; so substituting for X,Y,Z I get D1 to be

[(x1^2-x2^2+y1^2-y2^2+z1^2-z2^2]/2=D1

Similarly the plane perpendicular to 2-3 containing the midpoint of 2-3 is

eq3)(x2-x3)X+(y2-y3)Y+(z2-z3)=D2

and

D2=[x2^-x3^2+y2^2-y3^2+z2^2-z3^2]/2

So the problem boils down to solving the 3x3 for for a,b,c, then evaluating D1 and D2 and finally solving for X.Y, Z from the 3x3 set of the 3 planes, which is formally more concise.

But I have no problem for those who need more clarity and don't feel comfortable working in space. and as far a time to do it, it's not that much greater, but why all the additional steps.

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

Security Warning Macros have been disabled [Options]

You need to click on the Options box and select "Trust this content"

In earlier versions the option to enable macros is much more obvious.

Pleae let me know what version you are using, and if enabling macros fixes the problem.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

The centre of the required bend radius lays somewhere along the bisector of the two tangent lines that create the point (B) of intersection of the tangents — and this distance, from point B, of this radius centre along this bisector, is the hypotenuse of a right-angled triangle (either ABD or CBD) — depending which is the shorter of the two lines. By finding the bisector of lines AB and BC between A and C, the direction cosines of the bisector are calculated, This distance from B to the radius centre is the length of the line from the closest given point (A or C) to the intersection point, divided by the cosine of half the angle ABC.

To avoid using the other of the two bisectors of the two lines, I've found the bisector line using the mid-point of two points spaced equal distances along the lines AB and BC (I've used an arbitary 1000 units, which is preferably longer than AB or BC.)

The position of the bend radius centre is found by multiplying the direction cosines of the bisector by the length of the hypotenuse - and adding them to the x,y,z values of the intersection point (B).

As before, it will run in an Excel VBA macro: I've tried to keep the thing 'readable' . I haven't (yet) done a check for the three point co-linear case, although it will flip through the straight-line situation. Once again, I've had someone check it against a 3D model — and it works — whether the next pipe bend is a LH or RH one. It also outputs identical results to Doug's method.

## CODE

Range("A1:Z512").NumberFormat = "###0.000000"

' ———————————————————————————————————————————————————————————————————

' Given three points in space (A,B,C)

Xa = -71: Ya = -37: Za = 47: ' Start Point

Xb = 59: Yb = -40: Zb = 0 ' Tangent Intersection Point

Xc = 100: Yc = 7: Zc = 88: ' Finish Point

' Lengths of AB, AC, AC

AB = (((Xa - Xb) ^ 2) + ((Ya - Yb) ^ 2) + ((Za - Zb) ^ 2)) ^ 0.5

BC = (((Xb - Xc) ^ 2) + ((Yb - Yc) ^ 2) + ((Zb - Zc) ^ 2)) ^ 0.5

AC = (((Xa - Xc) ^ 2) + ((Ya - Yc) ^ 2) + ((Za - Zc) ^ 2)) ^ 0.5

' Direction cosines of AB

ABi = (Xa - Xb) / AB

ABj = (Ya - Yb) / AB

ABk = (Za - Zb) / AB

' Direction cosines of BC

BCi = (Xc - Xb) / BC

BCj = (Yc - Yb) / BC

BCk = (Zc - Zb) / BC

' From point B create two points (G and H) 1000 units along AB and BC

Xg = Xb + (1000 * ABi)

Yg = Yb + (1000 * ABj)

Zg = Zb + (1000 * ABk)

'

Xh = Xb + (1000 * BCi)

Yh = Yb + (1000 * BCj)

Zh = Zb + (1000 * BCk)

' Find mid-point (M) of G and H

Xm = (Xg + Xh) / 2

Ym = (Yg + Yh) / 2

Zm = (Zg + Zh) / 2

' Length of line BM (Tangents bisector)

BM = ((Xm - Xb) ^ 2 + (Ym - Yb) ^ 2 + (Zm - Zb) ^ 2) ^ 0.5

' Direction cosines of BM

BMi = (Xm - Xb) / BM

BMj = (Ym - Yb) / BM

BMk = (Zm - Zb) / BM

' Cosine of angle ABM (Half angle ABC)

cosABM = (ABi * BMi) + (ABj * BMj) + (ABk * BMk)

' Find the shorter of AB and BC

If AB < BC Then Base = AB Else Base = BC

' Right-angled triange ABD or CBD

Hypotenuse = Base / cosABM

' Find coordinates of Bend Radius Centre

X_centre = Xb + (Hypotenuse * BMi)

Y_centre = Yb + (Hypotenuse * BMj)

Z_centre = Zb + (Hypotenuse * BMk)

' Find Bend Radius

Bend_Rad = (Hypotenuse ^ 2 - Base ^ 2) ^ 0.5

' Results:—

Cells(1, 1) = "Bend Radius": Cells(1, 2) = Bend_Rad

Cells(2, 1) = "X_centre ": Cells(2, 2) = X_centre

Cells(3, 1) = "Y_centre ": Cells(3, 2) = Y_centre

Cells(4, 1) = "Z_centre ": Cells(4, 2) = Z_centre

Cells(6, 1) = "Given Points"

Cells(7, 1) = "Point A(x,y,z)": Cells(7, 2) = Xa

Cells(7, 3) = Ya: Cells(7, 4) = Za

Cells(8, 1) = "Point B(x,y,z)": Cells(8, 2) = Xb

Cells(8, 3) = Yb: Cells(8, 4) = Zb

Cells(9, 1) = "Point C(x,y,z)": Cells(9, 2) = Xc

Cells(9, 3) = Yc: Cells(9, 4) = Zc

End Sub

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

and the co-ords of A and C,

you know the line thru A (and C) normal to AB (and BC)

these two lines intersect at the centre (O),

and the radius is the distance OC.

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

More than half the code in the new function is to check that the two tangent points are the same distance from the IP, and if necessary moving the further one so that they are.

The new function runs about 7 times quicker than the earlier one.

## CODE

' Find the centre and radius of a circular arc defined by two tangent points

' and the intersection point of the tangents

Dim DL1(1 To 4) As Double, DL2(1 To 4) As Double, DL3(1 To 3) As Double, DL4(1 To 3) As Double

Dim MP(1 To 3) As Double, a As Double, b As Double, c As Double, d As Double

Dim NDims As Long, i As Long

Dim Arccen(1 To 2, 1 To 4) As Double

Const MaxDiff As Double = 0.0000000001

If TypeName(TP1) = "Range" Then TP1 = TP1.Value2

If TypeName(TP2) = "Range" Then TP2 = TP2.Value2

If TypeName(IntP) = "Range" Then IntP = IntP.Value2

NDims = UBound(TP1, 2)

' Check that tangent points are equidistant from IntP

For i = 1 To NDims

DL1(i) = IntP(1, i) - TP1(1, i)

DL2(i) = IntP(1, i) - TP2(1, i)

Next i

DL1(4) = (DL1(1) ^ 2 + DL1(2) ^ 2 + DL1(3) ^ 2) ^ 0.5

DL2(4) = (DL2(1) ^ 2 + DL2(2) ^ 2 + DL2(3) ^ 2) ^ 0.5

' If not, scale the longer line towards the IP

If Abs(DL2(4) / DL1(4) - 1) > MaxDiff Then

If DL2(4) > DL1(4) Then

ScaleFact = DL1(4) / DL2(4)

For i = 1 To NDims

DL2(i) = DL2(i) * ScaleFact

TP2(1, i) = IntP(1, i) - DL2(i)

Arccen(2, i) = TP2(1, i)

Next i

DL2(i) = DL2(i) * ScaleFact

Arccen(2, 4) = 2

Else

ScaleFact = DL2(4) / DL1(4)

For i = 1 To NDims

DL1(i) = DL1(i) * ScaleFact

TP1(1, i) = IntP(1, i) - DL1(i)

Arccen(2, i) = TP1(1, i)

Next i

DL1(i) = DL1(i) * ScaleFact

Arccen(2, 4) = 1

End If

End If

' Point A = TP1, B = IP, C = mid-point of TP1-TP2

'Find side lengths of triangle ABC, a, b, c

'Find coordinates of Point C, midpoint of TP1-TP2

For i = 1 To NDims

MP(i) = (TP1(1, i) + TP2(1, i)) / 2

DL3(i) = (MP(i) - IntP(1, i))

DL4(i) = (MP(i) - TP1(1, i))

Next i

a = (DL3(1) ^ 2 + DL3(2) ^ 2 + DL3(3) ^ 2) ^ 0.5

b = (DL4(1) ^ 2 + DL4(2) ^ 2 + DL4(3) ^ 2) ^ 0.5

c = DL1(4)

' Radius = bc / a

' d = (distance arc centre to IP)/a = Rc / ba

R = b * c / a

d = R * c / (b * a)

For i = 1 To NDims

Arccen(1, i) = IntP(1, i) + DL3(i) * d

Next i

Arccen(1, 4) = R

ArcCenT2IP2 = Arccen

End Function

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

well that's pretty redundant, 'cause they have to be, congruent triangles.

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

If both the points are tangent points to the same circle then they will be eqidistant from the intersection point, but it's perfectly possible to enter points that are not equidistant from the intersection point, in which case the function would return the centre of a circle which was not tangential to either point, and you wouldn't have any idea about the mistake. By checking that the points are equidistant it picks up that potential error, and by reporting the coordinates for the adjusted point it gives an indication of the error.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

Here is the simplest hand calc solution according to me:

1. |C^-X1^| = |C^-X2^| = |C^-X3^|

2. C must lie on plane described by X1, X2 and X3.

The above 2 conditions describe a problem with 3 equations and 3 unknowns.

Where

C^, X1^, X2^ and X3^ are the position vectors of centers and the 3 points respectively. C = [c1 c2 c3], X1 = [x11 x12 x13] ... and so on.

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

By assuming that one point is a true tangent point and the other point lies on a tangent, but is not on the circle. From examining the coordinates of the adjusted point (which are in the second row of the function return array) it will be obvious whether this is a reasonable assumption or not.

Depends on the application. It would be very easy to modify the code to do that. If you did you'd want to allow the passing of an acceptable error value (which would obviously depend on the units). For the application being discussed in this thread, which uses physical meaurements of a bent pipe, the accuracy would almost never match that required by the function, so it would almost always return an error, which wouldn't be very useful. You could easily re-write it to allow an error of up to some realistic value, with an adjustment of the further tangent point, and to return an error if the discrepancy in the tangent lengths was greater than that specified.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

- On the error issue ... this is actually what we are dealing with. We are given the 3 pts by an outside Engr party. When checking first for both tangent lines being equal, they are not equal (in all 4 cases for us). What we did was to move the point of intersection (POI) in the direction that had the most effect on the length, and the least affect on the positon of the pipe in space. So the small errors are corrected mathematically very quickly, but with little physical effect on the structure. We are dealing very large diameters (22'R to 580'R), and relatively small circle arc lengths (only 6'L to 20'L). We found that some very slight movements or errors in the 3 point's coordinate positions can have very large affects on the bend raddii. For example, if the POI "height" above the arc changes very slightly (e.g. 1"), it can change the arc radius by several feet. This is because it is a very "flat" and short arc. For example, taking this to extremes, if you had a 10' arc length on a 10,000' radius, the POI almost rests right on the circle. Raise that POI by just 1" and see how mach that radius has to change to accomodate that. Again, thanks for all this great help. Mike P.

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

with your dimensions i would've thought the problem was in accurately measuring the dimensions, i would've expected at a 3D CAD program (or even a surveying program) would have been able to accurately dimension the required geometry. i suspect the real problem is how to physically create the required shape, rather than how to mathematically define it.

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

- I agree .. I think we have beat this one to death. But to answer your question ... we have to define it before we can make it ... and making this shape is not a big problem ... we have bent the 1st two short pipes and can make them lay on a true cut template within 1/16". Accuracy is important in our case because these pipes are not just out in free space. They are attached to a heavy fabicated cradle system, with 2"T support plates every 24". The support plates have half-circle cutouts that "cradle" the pipes. The support structure with the cradle plates are all fabricated as an assembly first. Then the bent pipe has to lay into this 3D structure. So we need to do both ... DEFINE the curve, and then PRODUCE the curve to a fair degree of accuracy for all of this to come together properly. I could have said this early on, but didn't want to cloud the issue any more than it was. Thanks again, Mike P.

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

More details and download from:

http://

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

=====================================

Eng-tips forums: The best place on the web for engineering discussions.

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

- Thanks again for postings. Most helpful.

- By any chance do you have a couple of related formulas?

1) After determining the circle radius and center point, do you have an Excel solution to determine the coordinates of any other point on the circle given any one of the X-Y-Z coordinates of the unknown point, OR given the arc angle from another known point on the circle?

2) Knowing the coordinates of 2 points in 3D (i.e. a line in 3D), an Excel method to determine the coordinates of any other point on the line by just plugging in a distance along the line from one of the two known points? This is pretty simple with the distance formula. If you don't already have this written, I can write it. I thought I'd check first to see if you already have one done in Excel.

Thanks again,

Mike P.

## RE: Find circle radius & circle center in 3D from (3) XYZ pts

My plan is to split out the interpolation part of the IP spreadsheet, and add in additional geometric functions and line plotting functions, and make it a complete geometric tool-box, but I don't know when I'll get round to doing that.

Doug Jenkins

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