Given point a point a (x1,y1) and a point c (x3,y3) we can calculate a slope m. Assuming we have a distance d I'm a bit stuck trying to figure out how to find a point b (x2,y2) which is d distance from x1,y1 in the direction of c.
Does anyone know how to calculate this? I thought about using the midpoint function but it's not quite there.
Help?
You can work out the full distance between a and c with:
__________________________________
df = / (x3-x1)*(x3-x1) + (y3-y1)*(y3-y1)
\/
This uses the standard "root of the sum of squares" method.
Then, if the actual partial distance you want is dp, the point can be found at (x2,y2) with:
x2 = x1 + dp/df * (x3-x1)
y2 = y1 + dp/df * (y3-y1)
which is simply moving the correct proportion dp/df in both dimensions.
You can get the direction from A to B by the following:
D = B - A
Then, you may normalize the direction (which means it is magnitude 1, or length 1):
N = D / D.Length
where
D.Length = sqrt(D.X * D.X + D.Y * D.Y)
To find a point on the line given by A and B, X units away from A in the direction of B, you would use the following:
Final = A + N * X
Related
I'm currently trying to write a program to determine if a line is going through the area of a square of 4 points and I'm searching for a formula. I only found solutions for 3-dimensional planes with vectors and tried to apply them to my situation by calculating with pen and paper but I always seem to hit a dead end when a third value is being needed.
I think the best way to approach it is to calculate the distance of the line to the square. Which would be 0 if it is passing through (a part of it) obviously. But I can't seem to find the right words for the google and stack overflow search since this seems too basic to not have been answered before.
If anyone has a link or a suggestion on how to calculate this I would be really thankful.
For my formula testing I've been working with these simple values:
Line:
l = (0 , 0) + s * (10, 10)
Points of the Square:
A (5, 5)
B (6, 5)
C (6, 6)
D (5, 6)
EDIT:
Using the function of the reply I marked as the answer I got it to work. A problem I had was getting the correct input for the function. The variables a, b and c. This is how I got them in the end:
var a = 1 / x2
var b = -(1 / y2)
var c = y1/y2 - x1/x2
Here's an idea on how you can approach the this problem.
First, what does line passing through a square means in a coordinate system?
Line L passes through square ABCD if and only if L separates the diagonally opposite sides on ABCD (A&C or B&D). Now the problem simplifies to checking whether two given points are separated by a given line.
Let the equation of the line L be ax + by + c = 0. Define a function f(x,y) = ax + by + c. Point A=(x1,y1) and C=(x2,y2) are separated by line L if f(x1,y1) and f(x2,y2) have opposite signs. Additionally if they have the same sign it means for point are on the same side of the line.
Here's the Python code for the above idea:
# Function to check if two points
# lie on the opposite side of the line
def pointsAreOnOppositeSideOfLine(a, b, c, x1, y1, x2, y2):
fx1 = 0 # Variable to store a * x1 + b * y1 - c
fx2 = 0 # Variable to store a * x2 + b * y2 - c
fx1 = a * x1 + b * y1 - c
fx2 = a * x2 + b * y2 - c
# If fx1 and fx2 have same sign
if ((fx1 * fx2) <= 0):
return True
return False
I have a line AB. I would like to draw a line BC, perpendicular to AB. I know xyz of the points A and B, I also know the distance N between B and C. How can I find an arbitrary point C which fits into the given parameters? The calculations should be done in 3-D. Any point, perpendicular to AB can be the point C, if its distance to B equals N.
An almost identical question is given here, but I would like to know how the same thing is done in 3-D: How do you find a point at a given perpendicular distance from a line?
The calculation that works for me in 2-D was given in the link above:
dx = A.x-B.x
dy = A.y-B.y
dist = sqrt(dx*dx + dy*dy)
dx /= dist
dy /= dist
C.x = B.x + N*dy
C.y = B.y - N*dx
I tried adding Z axis to it like this:
dz = A.z - B.z
dist = sqrt(dx*dx + dy*dy + dz*dz)
dz /=dist
C.z = .... at this point it becomes a mystery to me
If I put something like "C.z - N*dz" into C.z, the distance is accurate only in some rotation angles, I would like to know the correct solution. I can imagine that in 3-D it is calculated in a completely different manner.
Clarification
Point C is not unique. It can be any point on a circle with its
centre at B and radius N. The circle is perpendicular to AB
If the desired point C can be any of the infinitely-many points fitting your requirements, here is one method.
Choose any vector that is not parallel or anti-parallel to vector AB. You could try the vector (1, 0, 0), and if that is parallel you could use (0, 1, 0) instead. Then take the cross-product of vector AB and the chosen vector. That cross-product is perpendicular to vector AB. Divide that cross-product by its length then multiply by the desired length N. Finally extend that vector from point B to find your desired point C.
Here is code in Python 3 that follows that algorithm. This code is somewhat non-pythonic to make it easier to convert to other languages. (If I really did this for myself I would use the numpy module to avoid coordinates completely and shorten this code.) But it does treat the points as tuples of 3 values: many languages will require you to handle each coordinate separately. Any real-life code would need to check for "near zero" rather than "zero" and to check that the sqrt calculation does not result in zero. I'll leave those additional steps to you. Ask if you have more questions.
from math import sqrt
def pt_at_given_distance_from_line_segment_and_endpoint(a, b, dist):
"""Return a point c such that line segment bc is perpendicular to
line segment ab and segment bc has length dist.
a and b are tuples of length 3, dist is a positive float.
"""
vec_ab = (b[0]-a[0], b[1]-a[1], b[2]-a[2])
# Find a vector not parallel or antiparallel to vector ab
if vec_ab[1] != 0 or vec_ab[2] != 0:
vec = (1, 0, 0)
else:
vec = (0, 1, 0)
# Find the cross product of the vectors
cross = (vec_ab[1] * vec[2] - vec_ab[2] * vec[1],
vec_ab[2] * vec[0] - vec_ab[0] * vec[2],
vec_ab[0] * vec[1] - vec_ab[1] * vec[0])
# Find the vector in the same direction with length dist
factor = dist / sqrt(cross[0]**2 + cross[1]**2 + cross[2]**2)
newvec = (factor * cross[0], factor * cross[1], factor * cross[2])
# Find point c such that vector bc is that vector
c = (b[0] + newvec[0], b[1] + newvec[1], b[2] + newvec[2])
# Done!
return c
The resulting output from the command
print(pt_at_given_distance_from_line_segment_and_endpoint((1, 2, 3), (4, 5, 6), 2))
is
(4.0, 6.414213562373095, 4.585786437626905)
Hello again first part is working like a charm, thank you everyone.
But I've another question...
As I've no interface, is there a way to do the same thing with out not knowing the radius of the circle?
Should have refresh the page CodeMonkey solution is exactly what I was looking for...
Thank you again.
============================
First I'm not a developer, I'm a simple woodworker that left school far too early...
I'm trying to make one of my tool to work with an autonomous robot.
I made them communicate by reading a lot of tutorials.
But I have one problem I cant figure out.
Robot expect position of the tool as (X,Y) but tool's output is (A,B,C)
A is the distance from tool to north
B distance to east
C distance at 120 degree clockwise from east axe
the border is a circle, radius may change, and may or may not be something I know.
I've been on that for 1 month, and I can't find a way to transform those value into the position.
I made a test with 3 nails on a circle I draw on wood, and if I have the distance there is only one position possible, so I guess its possible.
But how?
Also, if someone as an answer I'd love pseudo code not code so I can practice.
If there is a tool to make a drawing I can use to make it clearer can you point it out to me?
Thank you.
hope it helps :
X, Y are coordinate from center, Da,Db, Dc are known.
Trying to make it more clear (sorry its so clear in my head).
X,Y are the coordinate of the point where is the tool (P).
Center is at 0,0
A is the point where vertical line cut the circle from P, with Da distance P to A;
B is the point where horizontal line cuts the circle fom P, with Db distance P to B.
C is the point where the line at 120 clockwise from horizontal cuts the circle from P, with Dc distance P to C.
Output from tool is an array of int (unit mm): A=123, B=114, C=89
Those are the only informations I have
thanks for all the ideas I'll try them at home later,
Hope it works :)
Basic geometry. I decided to give up having the circle at the origin. We don't know the center of the circle yet. What you do have, is three points on that circle. Let's try having the tool's position, given as P, as the new (0,0). This thus resolves to finding a circle given three points: (0, Da); (Db,0), and back off at 120° at Dc distance.
Pseudocode:
Calculate a line from A to B: we'll call it AB. Find AB's halfway point. Calculate a line perpendicular to AB, through that midpoint (e.g. the cross product of AB and a unit Z axis finds the perpendicular vector).
Calculate a line from B to C (or C to A works just as well): we'll call it BC. Find BC's halfway point. Calculate a line perpendicular to BC, through that midpoint.
Calculate where these two lines cross. This will be the origin of your circle.
Since P is at (0,0), the negative of your circle's origin will be your tool's coordinates relative to the circle's origin. You should be able to calculate anything you need relative to that, now.
Midpoint between two points: X=(X1+X2)/2. Y=(Y1+Y2)/2.
The circle's radius can be calculated using, e.g. point A and the circle's origin: R=sqrt(sqr((Ax-CirX)+sqr(Ay-CirY))
Distance from the edge: circle's radius - tool's distance from the circle's center via Pythagorean Theorem again.
Assume you know X and Y. R is the radius of the circle.
|(X, Y + Da)| = R
|(X + Db, Y)| = R
|(X - cos(pi/3) * Dc, Y - cos(pi/6) * Dc)| = R
Assuming we don't know the radius R. We can still say
|(X, Y + Da)|^2 = |(X + Db, Y)|^2
=> X^2 + (Y+Da)^2 = (X+Db)^2 + Y^2
=> 2YDa + Da^2 = 2XDb + Db^2 (I)
and denoting cos(pi/3)*Dc as c1 and cos(pi/6)*Dc as c2:
|(X, Y + Da)|^2 = |(X - c1, Y - c2)|^2
=> X^2 + Y^2 + 2YDa + Da^2 = X^2 - 2Xc1 + c1^2 + Y^2 - 2Yc2 + c2^2
=> 2YDa + Da^2 = - 2Xc1 + c1^2 - 2Yc2 + c2^2
=> Y = (-2Xc1 + c1^2 + c2^2 - Da^2) / 2(c2+Da) (II)
Putting (II) back in the equation (I) we get:
=> (-2Xc1 + c1^2 + c2^2 - Da^2) Da / (c2+Da) + Da^2 = 2XDb + Db^2
=> (-2Xc1 + c1^2 + c2^2 - Da^2) Da + Da^2 * (c2+Da) = 2XDb(c2+Da) + Db^2 * (c2+Da)
=> (-2Xc1 + c1^2 + c2^2) Da + Da^2 * c2 = 2XDb(c2+Da) + Db^2 * (c2+Da)
=> X = ((c1^2 + c2^2) Da + Da^2 * c2 - Db^2 * (c2+Da)) / (2Dbc2 + 2Db*Da + 2Dac1) (III)
Knowing X you can get Y by calculating (II).
You can also make some simplifications, e.g. c1^2 + c2^2 = Dc^2
Putting this into Python (almost Pseudocode):
import math
def GetXYR(Da, Db, Dc):
c1 = math.cos(math.pi/3) * Dc
c2 = math.cos(math.pi/6) * Dc
X = ((c1**2 + c2**2) * Da + Da**2 * c2 - Db * Db * (c2 + Da)) / (2 * Db * c2 + 2 * Db * Da + 2 * Da * c1)
Y = (-2*X*c1 + c1**2 + c2**2 - Da**2) / (2*(c2+Da))
R = math.sqrt(X**2 + (Y+Da)**2)
R2 = math.sqrt(Y**2 + (X+Db)**2)
R3 = math.sqrt((X - math.cos(math.pi/3) * Dc)**2 + (Y - math.cos(math.pi/6) * Dc)**2)
return (X, Y, R, R2, R3)
(X, Y, R, R2, R3) = GetXYR(123.0, 114.0, 89.0)
print((X, Y, R, R2, R3))
I get the result (X, Y, R, R2, R3) = (-8.129166703588021, -16.205081335032794, 107.1038654949096, 107.10386549490958, 107.1038654949096)
Which seems reasonable if both Da and Db are longer than Dc, then both coordinates are probably negative.
I calculated the Radius from three equations to cross check whether my calculation makes sense. It seems to fulfill all three equations we set up in the beginning.
Your problem is know a "circumscribed circle". You have a triangle define by 3 distances at given angles from your robot position, then you can construct the circumscribed circle from these three points (see Circumscribed circle from Wikipedia - section "Other properties"). So you know the diameter (if needed).
It is also known that the meeting point of perpendicular bisector of triangle sides is the center of the circumscribed circle.
Let's a=Da, b=Db. The we can write a system for points A and B at the circumference:
(x+b)^2 + y^2 = r^2
(y+a)^2 + x^2 = r^2
After transformations we have quadratic equation
y^2 * (4*b^2+4*a^2) + y * (4*a^3+4*a*b^2) + b^4-4*b^2*r^2+a^4+2*a^2*b^2 = 0
or
AA * y^2 + BB * y + CC = 0
where coefficients are
AA = (4*b^2+4*a^2)
BB = (4*a^3+4*a*b^2)
CC = b^4-4*b^2*r^2+a^4+2*a^2*b^2
So calculate AA, BB, CC coefficients, find solutions y1,y2 of quadratic eqiation, then get corresponding x1, x2 values using
x = (a^2 - b^2 + 2 * a * y) / (2 * b)
and choose real solution pair (where coordinate is inside the circle)
Quick checking:
a=1,b=1,r=1 gives coordinates 0,0, as expected (and false 1,-1 outside the circle)
a=3,b=4,r=5 gives coordinates (rough) 0.65, 1.96 at the picture, distances are about 3 and 4.
Delphi code (does not check all possible errors) outputs x: 0.5981 y: 1.9641
var
a, b, r, a2, b2: Double;
aa, bb, cc, dis, y1, y2, x1, x2: Double;
begin
a := 3;
b := 4;
r := 5;
a2 := a * a;
b2:= b * b;
aa := 4 * (b2 + a2);
bb := 4 * a * (a2 + b2);
cc := b2 * b2 - 4 * b2 * r * r + a2 * a2 + 2 * a2 * b2;
dis := bb * bb - 4 * aa * cc;
if Dis < 0 then begin
ShowMessage('no solutions');
Exit;
end;
y1 := (- bb - Sqrt(Dis)) / (2 * aa);
y2 := (- bb + Sqrt(Dis)) / (2 * aa);
x1 := (a2 - b2 + 2 * a * y1) / (2 * b);
x2 := (a2 - b2 + 2 * a * y2) / (2 * b);
if x1 * x1 + y1 * y1 <= r * r then
Memo1.Lines.Add(Format('x: %6.4f y: %6.4f', [x1, y1]))
else
if x2 * x2 + y2 * y2 <= r * r then
Memo1.Lines.Add(Format('x: %6.4f y: %6.4f', [x2, y2]));
From your diagram you have point P that you need it's X & Y coordinate. So we need to find Px and Py or (Px,Py). We know that Ax = Px and By = Py. We can use these for substitution if needed. We know that C & P create a line and all lines have slope in the form of y = mx + b. Where the slope is m and the y intercept is b. We don't know m or b at this point but they can be found. We know that the angle of between two vectors where the vectors are CP and PB gives an angle of 120°, but this does not put the angle in standard position since this is a CW rotation. When working with circles and trig functions along with linear equations of slope within them it is best to work in standard form. So if this line of y = mx + b where the points C & P belong to it the angle above the horizontal line that is parallel to the horizontal axis that is made by the points P & B will be 180° - 120° = 60° We also know that the cos angle between two vectors is also equal to the dot product of those vectors divided by the product of their magnitudes.
We don't have exact numbers yet but we can construct a formula: Since theta = 60° above the horizontal in the standard position we know that the slope m is also the tangent of that angle; so the slope of this line is tan(60°). So let's go back to our linear equation y = tan(60°)x + b. Since b is the y intercept we need to find what x is when y is equal to 0. Since we still have three undefined variables y, x, and b we can use the points on this line to help us here. We know that the points C & P are on this line. So this vector of y = tan(60°)x + b is constructed from (Px, Py) - (Cx, Cy). The vector is then (Px-Cx, Py-Cy) that has an angle of 60° above the horizontal that is parallel to the horizontal axis. We need to use another form of the linear equation that involves the points and the slope this time which happens to be y - y1 = m(x - x1) so this then becomes y - Py = tan(60°)(x - Px) well I did say earlier that we could substitute so let's go ahead and do that: y - By = tan(60°)(x - Ax) then y - By = tan(60°)x - tan(60°)Ax. And this becomes known if you know the actual coordinate points of A & B. The only thing here is that you have to convert your angle of 120° to standard form. It all depends on what your known and unknowns are. So if you need P and you have both A & B are known from your diagram the work is easy because the points you need for P will be P(Ax,By). And since you already said that you know Da, Db & Dc with their lengths then its just a matter of apply the correct trig functions with the proper angle and or using the Pythagorean Theorem to find the length of another leg of the triangle. It shouldn't be all that hard to find what P(x,y) is from the other points. You can use the trig functions, linear equations, the Pythagorean theorem, vector calculations etc. If you can find the equation of the line that points C & P knowing that P has A's x value and has B's y value and having the slope of that line that is defined by the tangent above the horizontal which is 180° - phi where phi is the angle you are giving that is CW rotation and theta would be the angle in standard position or above the horizontal you have a general form of y - By = tan(180° - phi)(x - Ax) and from this equation you can find any point on that line.
There are other methods such as using the existing points and the vectors that they create between each other and then generate an equilateral triangle using those points and then from that equilateral if you can generate one, you can use the perpendicular bisectors of that triangle to find the centroid of that triangle. That is another method that can be done. The only thing you may have to consider is the linear translation of the line from the origin. Thus you will have a shift in the line of (Ax - origin, By - origin) and to find one set the other to 0 and vise versa. There are many different methods to find it.
I just showed you several mathematical techniques that can help you to find a general equation based on your known(s) and unknown(s). It just a matter of recognizing which equations work in which scenario. Once you recognize the correct equations for the givens; the rest is fairly easy. I hope this helps you.
EDIT
I did forget to mention one thing; and that is the line of CP has a point on the edge of the circle defined by (cos(60°), sin(60°)) in the 1st quadrant. In the third quadrant you will have a point on this line and the circle defined by (-cos(60°), -sin(60°)) provided that this line goes through the origin (0,0) where there is no y nor x intercepts and if this is the case then the point on the circle at either end and the origin will be the radius of that circle.
So I was reading over something on this page (http://gamedeveloperjourney.blogspot.com/2009/04/point-plane-collision-detection.html)
The author mentioned
d = - D3DXVec3Dot(&vP1, &vNormal);
where vP1 is a point on the plane and vNormal is the normal to the plane. I'm curious as to how this gets you the distance from the world origin since the result will always be 0. In addition, just to be clear (since I'm still kind of hazy on the d part of a plane equation), is d in a plane equation the distance from a line through the world origin to the plane's origin?
In the generic case the distance between a point p and a plane can be computed by
<p - p0, normal>
where <a, b> is the dot product operation
<a, b> = ax*bx + ay*by + az*bz
and where p0 is a point on the plane.
When n is of unity length the dot product between a vector and it is the (signed) length of the projection of the vector on the normal
The formula you are reporting is just the special case when the point p is the origin. In this case
distance = <origin - p0, normal> = - <p0, normal>
This equality is formally wrong because the dot product is about vectors, not points... but still holds numerically. Writing down the explicit formula you get that
(0 - p0.x)*n.x + (0 - p0.y)*n.y + (0 - p0.z)*n.z
is the same as
- (p0.x*n.x + p0.y*n.y + p0.z*n.z)
Indeed a nice way to store a plane is to save the normal n and the value of k = <p0, n> where p0 is any point on the plane (the value of k is independent on which point you choose of the plane).
The result is not always zero. The result will only be zero if the plane goes through the origin. (Here let's assume the plane doesn't go through the origin.)
Basically, you are given a line from the origin to some point on the plane. (I.e. you have a vector from the origin to vP1). The problem with this vector is that most likely it's slanted and going to some far away place on the plane rather than to the closest point on the plane. So, if you simply took the length of vP1 you will get a distance that is too big.
What you need to do is get the projection of vP1 onto some vector that you know is perpendicular to the plane. That of course is vNormal. So take the dot product of vP1 and vNormal, and divide by the length of vNormal and you have the answer. (If they are kind enough to give you a vNormal that already is magnitude one, then no need to divide.)
You can work this out with Lagrange multipliers:
You know that the closest point on the plane must be of the form:
c = p + v
Where c is the closest point and v is a vector along the plane (which is thus orthogonal to n, the normal). You are trying for find the c with the smallest norm (or norm squared). So you are trying to minimized dot(c,c) subject to v being orthogonal to n (thus dot(v,n) = 0).
Thus, set up Lagrangian:
L = dot(c,c) + lambda * ( dot(v,n) )
L = dot(p+v,p+v) + lambda * ( dot(v,n) )
L = dot(p,p) + 2*dot(p,v) + dot(v,v) * lambda * ( dot(v,n) )
And take the derivative with respect to v (and set to 0) to get:
2 * p + 2 * v + lambda * n = 0
You can solve for lambda by in the equation above by dot producting both sides by n to get
2 * dot(p,n) + 2 * dot(v,n) + lambda * dot(n,n) = 0
2 * dot(p,n) + lambda = 0
lambda = - 2 * dot(p,n)
Note again that dot(n,n) = 1 and dot(v,n) = 0 (since v is in the plane and n is orthogonal to it). Then subtitute lambda back in to get:
2 * p + 2 * v - 2 * dot(p,n) * n = 0
and solve for v to get:
v = dot(p,n) * n - p
Then plug this back into c = p + v to get:
c = dot(p,n) * n
The length of this vector is |dot(p,n)| and the sign tells you whether the point is in the direction of the normal vector from the origin, or the reverse direction from the origin.
I tried using a raycasting-style function to do it but can't get any maintainable results. I'm trying to calculate the intersection between two tangents on one circle. This picture should help explain:
I've googled + searched stackoverflow about this problem but can't find anything similar to this problem. Any help?
Well, if your variables are:
C = (cx, cy) - Circle center
A = (x1, y1) - Tangent point 1
B = (x2, y2) - Tangent point 2
The lines from the circle center to the two points A and B are CA = A - C and CB = B - C respectively.
You know that a tangent is perpendicular to the line from the center. In 2D, to get a line perpendicular to a vector (x, y) you just take (y, -x) (or (-y, x))
So your two (parametric) tangent lines are:
L1(u) = A + u * (CA.y, -CA.x)
= (A.x + u * CA.y, A.y - u * CA.x)
L2(v) = B + v * (CB.y, -CB.x)
= (B.x + v * CB.y, B.x - v * CB.x)
Then to calculate the intersection of two lines you just need to use standard intersection tests.
The answer by Peter Alexander assumes that you know the center of the circle, which is not obvious from your figure http://oi54.tinypic.com/e6y62f.jpg.
Here is a solution without knowing the center:
The point C (in your figure) is the intersection of the tangent at A(x, y) with the line L perpendicular to AB, cutting AB into halves. A parametric equation for the line L can be derived as follows:
The middle point of AB is M = ((x+x2)/2, (y+y2)/2), where B(x2, y2). The vector perpendicular to AB is N = (y2-y, x-x2). The vector equation of the line L is hence
L(t) = M + t N, where t is a real number.