I am trying to find intersecting Points for Circle with help of this link.
The following note describes how to find the intersection point(s) between two circles on a plane, the following notation is used. The aim is to find the two points P3 = (x3, y3) if they exist.
First calculate the distance d between the center of the circles. d = ||P1 - P0||.
If d > r0 + r1 then there are no solutions, the circles are separate.
If d < |r0 - r1| then there are no solutions because one circle is contained within the other.
If d = 0 and r0 = r1 then the circles are coincident and there are an infinite number of solutions.
Considering the two triangles P0P2P3 and P1P2P3 we can write
a2 + h2 = r02 and b2 + h2 = r12
Using d = a + b we can solve for a,
a = (r02 - r12 + d2 ) / (2 d)
It can be readily shown that this reduces to r0 when the two circles touch at one point, ie: d = r0 + r1
Solve for h by substituting a into the first equation, h2 = r02 - a2
So
P2 = P0 + a ( P1 - P0 ) / d
And finally, P3 = (x3,y3) in terms of P0 = (x0,y0), P1 = (x1,y1) and P2 = (x2,y2), is
x3 = x2 +- h ( y1 - y0 ) / d
y3 = y2 -+ h ( x1 - x0 ) / d
http://paulbourke.net/geometry/2circle/
b:=CircleMorph new.
b center: 60#60.
b openInWorld.
b1:=CircleMorph new.
b center: 100#100.
b1 openInWorld.
d:= b1 center - b center. // distance between 2 circles
r1:= (((b center x abs)squared +(b center y abs)squared)sqrt).
r2:= (((b1 center x abs)squared +(b1 center y abs)squared)sqrt).
r3:= r1+ r2.
(d) > (r3) ifTrue:[Transcript show:'Circles are seprate';cr]
When i compare distance with sum of radius of 2 circles is get distance less than radius of both circles which i know is not true,when circles are seprate Am i calculating radius correctly or not idk there is some problem with this help.
One possible solution is this:
| b b1 d r1 r2 r3 |
b := CircleMorph new.
b center: 60#60.
b bounds: (Rectangle origin: 100#40 corner: 40#100).
b openInWorld.
b1 := CircleMorph new.
b center: 100#100.
b bounds: (Rectangle origin: 100#40 corner: 40#100).
b1 openInWorld.
r1 := b bounds width / 2.
r2 := b1 bounds width / 2.
r3 := r1+ r2.
(d < r3)
ifTrue: [| a h mid |
a := (r1 squared - r2 squared + d squared) / (2 * d).
h := (r1 squared - a squared) sqrt.
mid := (b1 center x - b center x) # (b1 center y - b center y).
{(mid x + (h * (b1 center y - b center y))) # (mid y - (h * (b1 center x - b center x))).
(mid x - (h * (b1 center y - b center y))) # (mid y + (h * (b1 center x - b center x)))}]
ifFalse: ['separate']
You'll want to check my arithmetic. (Update: I hadn't used the point between the two circle centres to calculate the points.)
Related
I have two points A(x1,y1) & B(x2,y2) and I need to check if point c(x3,y3) falls on the straight line formed by point A & B.
A------C--------------------B then yes C is between A & B
A---------------------------B
C
in second case C isn't between A & B.
Using complex numbers, we define a similarity transformation that maps A to 0 and B to 1:
W = (Z - Za) / (Zb - Za)
Then Wc = (Zc - Za) / (Zb - Za) is a complex number that should have a zero or tiny imaginary value, and a real value between 0 and 1.
Make two vectors
cax = x1 - x3
cay = y1 - y3
cbx = x2 - x3
cby = y2 - y3
and check that angle between them is Pi:
Calculate dot and cross product of these vectors
dot = cax * cbx + cay * cby
cross = cax * cby - cay * cbx
Point C lies on segment AB if cross is zero and dot is negative
dot < 0
abs(cross) < eps
where eps is small value like 1e-6 to compensate floating point calculation errors.
if you are sure that all three point are aligned, you can use the dot product between vectors AC and AB, and the dot product between vectors BA and BC.
Reminder: Vector AB = (xB - xA, yB - yA)
Reminder: dot(AB, AC) = xABxAC + yAByAC
if dot(AB, AC) > 0, then it means C is in direction of B from A
if dot(BA, BC) > 0, then it means C is in direction of A from B
if both conditions above are satisfied, it means C is between A and B
I'm currently working on a program to do 4th order Runge Kutta calculations for ordinary differential equations., one of the requirements for the program is that the user will input the equation they want the 4th order operation to calculate. Is there a way to take the user inputted equation from a specific cell and use it in VBA sub to calculate the new y values?
The equations are going to be multi-variable polynomial equations containing x and y.
This is the loop I am hoping to use to perform the calculation, where equa is currently a function with a pre-established equation for testing, but is planned to be the user inputted equation.
n = (xf - xi) / h
For i = 1 To n
k1 = equa(x, y)
y1 = y + k1 * h / 2
k2 = equa(x + h / 2, y1)
y2 = y + k2 * h / 2
k3 = equa(x + h / 2, y2)
y3 = y + k3 * h / 2
k4 = equa(x + h, y3)
yf = y + ((k1 + 2 * k2 + 2 * k3 + k4) * (1 / 6) * h)
Cells(7 + i, 1).Value = y
x = x + h
Next i
Simple example using Evaluate:
Debug.Print Resolve("2*<x> + 3*<y>",1,2) '>>8
Function Resolve(sEq As String, x, y)
Resolve = Application.Evaluate(Replace(Replace(sEq, "<x>", x), "<y>", y))
End Function
I have points in 3D space.
X Y Z
0 0.61853 0.52390 0.26304
1 0.61843 0.52415 0.26297
2 0.62292 0.52552 0.26108
3 0.62681 0.51726 0.25622
4 0.62772 0.51610 0.25903
I have defined a plane through the points which should vertically divide these points, but it is not dividing them vertically or horizontally. The plane and the points are way apart while I'm plotting them.
def plane_equation(x1, y1, z1, x2, y2, z2, x3, y3, z3):
a1 = x2 - x1
b1 = y2 - y1
c1 = z2 - z1
a2 = x3 - x1
b2 = y3 - y1
c2 = z3 - z1
a = b1 * c2 - b2 * c1
b = a2 * c1 - a1 * c2
c = a1 * b2 - b1 * a2
d = (- a * x1 - b * y1 - c * z1)
return a, b, c, d
# Finding the equation of the plane
a, b, c, d = plane_equation(x0, y0, z0, x1, y1, z1, x2, y2, z2)
print("equation of plane is ", a, "x +", b, "y +", c, "z +", d, "= 0.")
x = np.arange(0, 1, 0.1)
y = np.arange(0, 1, 0.1)
X,Y = np.meshgrid(x,y)
Z = a*X + b*Y + d
fig = plt.figure()
ax = fig.gca(projection='3d')
ax.scatter(df.x, df.y, df.z, color = 'c', marker = 'o', alpha = 0.5)
surf = ax.plot_surface(X, Y, Z)
equation of plane is -0.0002496952000000007 x + 0.00036812320000000016 y + 0.0007697304000000002 z + -0.00024088567529317268 = 0.
I need plane to pass through these points and should be in vertical direction. The blue plane should pass through cyan points and plane should be in vertical direction.
There are several things that might go wrong here:
too small values
Your normal is not normalized and its coordinates are very small magnitude (0.000???) so its possible that your plot is handling all the values as zero (as the plot on your image is of the plane Z=0 which has nothing to do with values you provided).
From your feedback in chat This assumption of mine was right so to solve this problem just normalize your normal:
n(nx,ny,nz) /= sqrt(nx*nx + ny*ny +nz*nz)
And compute the d with this new values:
d = nx*x0 + ny*y0 + nz*z0
where (x0,y0,z0) is any from the selected points.
Wrongly selected points
The 3 selected points should be not too close to each other and not on a single line. If they do the computed normal is invalid. Also if you select points containing big noise the accuracy is lowered by it...
To improve this select 3 points randomly compute normal. Compute n such normals and average them together. The higher n the better accuracy.
Fit
To improve accuracy even more you can try to fit the normal and d. simply by using normal from #1 or #2 and fit its coordinates and d in near range to minimize the avg or max distance of all points to plane. However this is O(n.log^4(m)) where n is the number of used points and m relate to the fitted range of each parameter, but provides the best accuracy you can get.
You can use binary search or Approximation search or what ever optimizer your environment have at disposal
Given the following from the image below:
The green circle has a radius equal to B
The yellow line is tangent to the green circle
The vertical purple line is parallel to the green line and perpendicular to the yellow line.
The yellow line is perpendicular to both the green line and the vertical purple line
The purple point is centered on the green circle's edge
A and B are known values
I realize several of these constraints overlap, just trying to be thorough.
Pythagorean's theorem can provide the value of C, just to illustrate what I know we can determine already.
What is the formula/equation to determine D, where D is the perpendicular distance from the tangent yellow line to the arc/circle (at the purple point)?
Update
Replacing previous attempts to illustrate solution now with one that I can now visualize as the correct representation of the answer and comments provided by John
The distance D can be found by computing the lowest intersection between the vertical ray from the right endpoint of the yellow segment and the circle.
Some notations (x axis to the right, y axis to the bottom, origin at the center of the circle):
center of the circle: P_C = (0, 0)
origin of the vertical ray: P_O = (A, B)
direction of the vertical ray: v_d = (0, -1)
Points on the ray satisfy: P = P_O + t v_d = (A, B - t)
Points on the circle satisfy: |P P_O|^2 = B^2
Expanding the first equation into the second gives: A^2 + (B - t)^2 = B^2 = A^2 + B^2 - 2 B t + t^2
Solving t^2 - 2 B t + A^2 = 0 for t yields d = B^2 - A^2 > 0, so two solutions t_1 = B - sqrt(d), t_2 = B + sqrt(d) (one near the bottom of the circle, the other near the top as expected). But t actually gives the distance along the ray (since v_d is a unit vector), so what we are looking for is the smallest solution t_1. Hence D = B - sqrt(B^2 - A^2).
The final result can also be derived and / or verified geometrically (courtesy of John, see all the corresponding comments): D = B - B' and B'^2 + A^2 = B^2 (Pythagorus on the right triangle with the center of the circle and the purple point as two of its vertices and an edge sitting on the purple line).
As you mentioned C is the easy part. However with A,B,C and the cosine theorem you can work out the angel opposite to B (b):
cos(b) = (a^2 + c^2 -b^2)/(2ac)
knowing b and that A and D have a right angel you can work out the angle between C and D (b'):
b' = 90° - b
given that D lies on the circle you know that the distance from the center to D is B so you now have a triangle with sides B,D and C where you know two of the sides and one of the angles. With the cosine law again:
B^2 = C^2 + D^2 - 2CD cos(b')
so in one more step we can find:
B^2 - C^2 = D^2 - 2CD cos(b') + (C cos(b'))^2 -(C cos(b'))^2 <=>
B^2 - C^2 + (C cos(b'))^2 = (D - C cos(b'))^2 <=>
sqrt(B^2 - C^2 + (C cos(b'))^2) + C cos(b') = D
hope I didn't put stupid mistakes in there and this helps...
I have a square, for simplicity assume bottom left corner is on origin and width of the square is 1.
A ray divides the square into two parts. I have the coordinates of intersection points. I want to obtain the area that lies right of the vector from p1 to p2:
Right now I have 16 if statements checking every combination of 2 points and calculating the area accordingly. It looks awful. Is there a more clever way of doing this?
Call the points A and B instead of p1 and p2. I'll assume x increases to the right and y increases upward, as per convention.
The point A must have a coordinate (x or y) that is 0 or 1. Rotate the square (really just the two points) to make it x=0.
The point B might be at x=-1, in which case the area is 1-(Ay+By)/2.
Or B might be at y=0, area = 1+(AyBx)/2
Or B might be at y=1, area = (Ay-1)Bx/2
This solution assumes that p1 and p2 form a right-triangle as depicted in the shaded area:
Area to the right of the vector = (w * w) - (0.5 * p1 * p2)
where w is the width of the square, and 0 <= p1 <= w, and 0 <= p2 <= w.
For example if w = 1, p1 = 0.5, and p2 = 0.75
then Area = (1 * 1) - (0.5 * 0.5 * 0.75) = 0.8125