Urgent!!! Midpoint of Arc formula

in 2D

Anywhere on the line going through the mid point perpendicular to the line. (assuming homework I stop here)

You would need another point or a radius to fix it to a point.

Nah in 3D...

And i tried to look up extending a ray but no equations are available.

It is amazing how most softwares do it so easily.. the equations are hard to find.

the point lies on a polyhedron in the center of the sphere.

Definitely not homework.

What about using x² + y² + z² = R² ? but you would need R and a third point.

absolutely a homework of a middle school student.

you hve list sphere surface equation. and have two points. the midpoint coordinates is very evident.

With centre point and radius known

Mid point C =

xa + (xa - xb) / 2 , ya + (ya-xb) / 2 , za + (za-zb)/2

Point X can then be calculated by extending the vector Centre - C up to the radius.

I desperately need glasses. But I am standing because I cannot sit down for a long time.

You could go polar - determine Θ in the xy plane and then α in the z direction.

Xx = r Cos Θ

Xy = r sin Θ

Xz = r cos α

Here you go

http://williams.best.vwh.net/avform.htm

Write the parametric equation line thru "o"and "c",(at;bt;ct),look for the "t" for interjection line with surface sphere,put this value in the obtained line equation,now you got a point what belongs to the line and to the sphere (two points).-

The mid point of the arc cord is most easily given by adding the x's and dividing by two. Same for y's and z's. Then extend the ray from the origin through this point distance R.

That part of the problem reduces to solving two similar triangles which can be done using trigonometry or Pythagorius's theorem. I like Pythagorius. By substituting the mean coordinate point formula into the Pythagorian formula and working hard on the algebra you should be able to reduce the result to a single line that gives the R(x,y,z) of the tangent point of the spherical arc segment. I'm not interested enough in your problem to do the work for you.

First off the radius of the sphere is R = SQRT(x₁²+y₁²+z₁²) OR SQRT(x₂²+y₂²+z₂²)

Now the coordinates of the mid point of the line between the two points is

(x₃,y₃,z₃) = ((x₁+x₂)/2,(y₁+y₂)/2,(z₁+z₂)/2)

If the length of the line to the mid point is L = SQRT(x₃²+y₃²+z₃²)

Let the ratio of radius to the length to the mid point be k = R/L

Then the coords. of the mid point of the arc are (kx₃,ky₃,kz₃)

Have I missed the point here (no pun intended)?

The easiest method is, to use polar coordinates.

Assuming you use a coordinate system, which has its origin in the middle of the circle.

Then the equation of the circle is x^2+y^2=r^2

with x= r*cos (phi) and y = r*sin (phi)

you need to know only phi of the midpoint of the arc.

Use the coordiates of the two points, which you know and calculate their angles

alpha and beta

with xA = r*cos(alpha) and xB= r*cos(beta), you get alpha and beta, by using the inverse functions

alpha = arcus cosinus ( xA/r) and beta = arcus cosinus ( xB/r)

Subtract the angles and divide them by two:

(Alpha -Beta)/ 2 = phi

Put phi in the first two equations above and you have the coordinates of the mid point of the arc.

regards

Claus