I want to draw a triangle with only the 3 sides length. (In C# Winforms)
Example: S1(3), S2(4), S3(5) with SN(x) as the length of each side.
I've tried to do this and the result is not a right triangle but it looks like it.
For the first side i just draw it like a line. and after i triy to find the angle with Pythagore and the same for the second one, but i realised that if i enter (5,4,3) it's something else.I'm just try to understand how can i find coordinates of a triangle with only the length of the sides.
Point a = new Point(0, 0);
Point b = new Point(s1, 0);
double y = (Math.Pow(s1, 2) + Math.Pow(s3, 2) - Math.Pow(s2, 2)) / (2 * s1);
double x = Math.Sqrt(Math.Pow(s3, 2) - Math.Pow(y, 2));
Point c = new Point((int)x, (int)y);
e.Graphics.DrawLine(Pens.Black, a, b);
e.Graphics.DrawLine(Pens.Black, b, c);
e.Graphics.DrawLine(Pens.Black, c, a);
That's the result:
Can someone help me? because I think I don't understand how can I do this.
This is more a math problem. At point A you have the sides s1, s3 with opposing side s2. The cosine formula then gives
2*s1*s3*cos(alpha) = s1^2+s3^2-s2^2.
Now the cosine is the projection of the angle to the horizontal axis, so you should have
x = s3*cos(alpha) = (s1^2+s3^2-s2^2)/(2*s1)
and correspondingly
y = sqrt(s3^2-x^2).
For the test side lengths 3,4,5 this would give
x = (3^2 + (5^2-4^2))/(2*3) = 3
y = sqrt(5^2-3^2) = 4
producing the points for the rectangular triangle.
Related
Being not very good at math, I have a problem with my project.
The objective is boundary correction on 3D files.
In my application, the user moves a 3D point on X-axis in order to correct or modify the boundary of the object.
I want to move the nearest boundary points in the same direction but decreasingly. I mean no point should move more than the main point. The nearest points move most and, the farthest points should move less.
On the image, the red dots represent the initial status of points. And the user pulls the P0 in the x-direction. And the other points follow it. The last status of the points is represented by violet dots.
Here is what I tried.
//On point moved event
//Get nearest boundary Points (Uses Geometry3D to get boundary points).
(var clothDMesh, _) = Utilities3D.BuildDMesh(baseMesh);
CreateClothModel(clothDMesh);
var boundryVertices = nodes.Where(ro => ro.Value.isBorder).Select(ro => ro.Value.vertex).ToList();
var refPoint = CustomPoint.FromPoint3D(movedPoint);
//Gets the delta X.
var deltaX = p.X - initialX;
//Gets nearest country points, so 15 points above and 15 points below to move only a given number of points (I know maybe this should be calculated according to delta).
var nearestPoints = refPoint.GetNearesPoints(boundryVertices, 30);
foreach (var item in nearestPoints)
{
//This is only one of what I tried as a function. None of them worked correctly.
item.X += deltaX - (deltaX * 1/ Math.Pow(item.Distance, 2));
}
Any help will be appreciated.
Thanks in advance.
Here's the math part:
I call "a" your "deltaX".
We also need a second parameter: "b", the maximum height of the red dots. I assume it is symetrical and "-b" would be the minimum height of the red dots.
So, if you look for the value X, horizontal move, in fonction of the coordinate Y of the dot:
X = a - a * Y * Y / (b * b);
You can verify that for Y = 0, you obtain X = a and for Y = b (or -b) you get X = 0.
You have your parabola (X is function of Y^2).
I would like to draw a rectangle made of Mesh.
I enter the A starting point and B ending point.
The width of the rectangle is known in advance and is equal to H.
How to correctly determine the coordinates of corner points? (Everything happens in 3D)
There are a lot of theoretical entries on the net (but mostly for 2D) and trying something like this:
var vAB = B - A;
var P1 = new Vector3(-vAB.z, vAB.y, vAB.x) / Mathf.Sqrt(Mathf.Pow(vAB.x, 2) + Mathf.Pow(vAB.y, 2) + Mathf.Pow(vAB.z, 2)) * 0.5 * H;
But I can't find the correct solution
The simple way should be to use the cross product. The cross product of two vectors is perpendicular to both input vectors. You will need to define the normal of your rectangle, In this I use vector3.up. A-B cannot be parallel to the normal vector, or you will get an invalid result.
var l = B - A;
var s = Vector3.Normalize(Vector3.Cross(l, Vector3.up));
var p1 = A + s * h/2;
var p2 = A - s * h/2;
var p3 = B - s * h/2;
var p4 = B + s * h/2;
Here's a quick explanation of the trig involved. There are other tools which will reduce the boilerplate a bit, but this should give you an understanding of the underlying maths.
I've tweaked your problem statement slightly: I'm just showing the XY plane (there's no Z involved), and I've rotated it so that the line AB forms a positive angle with the horizontal (as it makes explaining the maths a bit easier). A is at (x1, y1), B is at (x2, y2).
The first step is to find the angle θ that the line AB makes with the horizontal. Draw a right-angled triangle, where AB is the hypotenuse, and the other two sides are parallel to the X and Y axes:
You can see that the horizontal side has length (x2 - x1), and the vertical side has length (y2 - y1). The angle between the base and the hypotenuse (the line AB) is given by trig, where tan θ = (y2 - y1) / (x2 - x1), so θ = arctan((y2 - y1) / (x2 - x1)).
(Make sure you use the Math.Atan2 method to calculate this angle, as it makes sure the sign of θ is correct).
Next we'll look at the corner P1, which is connected to A. As before, draw a triangle with the two shorter sides being parallel at the X and Y axes:
This again forms a right-angled triangle with hypotenuse H/2. To find the base of the triangle, which is the X-distance between P1 and A, again use trig: H/2 * sin θ. Similarly, the Y-distance between P1 and A is H/2 cos θ. Therefore P1 = (x1 + H/2 sin θ, y2 - H/2 cos θ).
Repeat the same trick for the other 3 corners, and you'll find the same result, but with differing signs.
My approach requires you to use a Transform.
public Transform ht; // assign in inspector or create new
void calculatePoints(Vector3 A, Vector3 B, float H)
{
Vector3 direction = B - A;
ht.position = A;
ht.rotation = Quaternion.LookRotation(direction, Vector3.Up);
Vector3 P1 = new Vector3(A + ht.right * H/2f);
Vector3 P2 = new Vector3(A + ht.left * H/2f);
Vector3 P3 = new Vector3(B + ht.right * H/2f);
Vector3 P4 = new Vector3(B + ht.left * H/2f);
}
I think it's intuitive to think with "left" and "right" which is why I used the Transform. If you wanted to define a point along the way, you'd be using ht.forward * value added to A, where value would be something between 0 and direction.magnitude.
I've already figured out how to make a 3D trajectory using a start point and an angle.
However, I am trying to make a trajectory from a start point, an end point, and a height.
I tried taking the approach of a parabola on a 2D plane in a 3D space. I calculated the Prabola's A, B, and C values as well as the plane it's on given 3 points on the Parabola.
However, I've had a few complications with this sort of calculation, I assume it has to do with the inability to properly calculate a Z-axis without a plane but I cannot tell.
Other than a 2D parabola on a plane google did not provide another possible answer and a 3D trajectory yields a formula using a start point, an angle, and a power multiplier.
Is there any way to calculate a 3D trajectory given the start point, end point, and height?
Appreciating your help
Edit:
My code to calculate a parabola using 3 points (in case someone would like to know how I've done that and perhaps fix what I've done wrong)
public Parabola(Vector3 pa, Vector3 pb, Vector3 pc)
{
this.pa = pa;
this.pc = pc;
float a1 = -pa.x * pa.x + pb.x * pb.x, b1 = -pa.x + pb.x, c1 = -pa.y + pb.y;
float a2 = -pb.x * pb.x + pc.x * pc.x, b2 = -pb.x + pc.x, c2 = -pb.y + pc.y;
float bm = -(b2 / b1), a3 = bm * a1 + a2, c3 = bm * c1 + c2;
float a = c3 / a3, b = (c1 - a1 * a) / b1, c = pa.y - a * pa.x * pa.x - b * pa.x;
this.a = a; this.b = b; this.c = c;
plane = Vector3.Cross(pb - pa, pc - pa);
}
public Vector3 GetPoint(float x)
{
float angle = Mathf.Atan2(pc.z - pa.z, pc.x - pa.x) * Mathf.Rad2Deg;
float xs = Mathf.Cos(angle * Mathf.Deg2Rad) * x, zs = Mathf.Sin(angle * Mathf.Deg2Rad) * x;
return new Vector3(xs, a * x * x + b * x + c, zs);
}
public Vector3 ProjectOn(float x) => Vector3.ProjectOnPlane(GetPoint(x), plane);
The result looks ok when it's only on 2 Axis, but not 3.
here are 2 images for demonstration:
Looking at the second image, the parabola seems to be correct aside from being scaled incorrectly. Let's take a look at your code.
public Vector3 GetPoint(float x)
{
float angle = Mathf.Atan2(pc.z - pa.z, pc.x - pa.x) * Mathf.Rad2Deg;
float xs = Mathf.Cos(angle * Mathf.Deg2Rad) * x, zs = Mathf.Sin(angle * Mathf.Deg2Rad) * x;
return new Vector3(xs, a * x * x + b * x + c, zs);
}
I'm making a lot of assumptions here, but it seems like x is meant as a value that goes from 0.0 to 1.0, representing the start and end of the parabola, respectively. If so, you are determining the X and Z coordinates of this point based exclusively on the sine/cosine of the angle and x. This means that the values xs and zs should only ever be able to be between -1 and 1, limiting yourself to the confines of the unit circle.
The values xs and zs look like they need to be scaled by a factor s calculated by measuring the 2D distance of the start and end points when projected onto the XZ plane. This should stretch the parabola just enough to reach the end point.
I found an answer, but it's kinda a workaround.
Before messing around with Parabolas in 3D, I messed around with linear equations in 3D.
Unlike parabolas, lines have a defined equation even in 3D(Pn = P0 + t x V)(Pn vector containing XYZ, P0 initial point containing XYZ, t float, V Vector3)
In addition, there's only ONE line that goes through 2 points, even in 3D.
I used that to make a trajectory that's made out of 2 points and a height.
I make a new point in the center of those two points and add the height value to the highest Y value of the points, thus creating an Apex.
then I use the same calculations as before to calculate the A, B, and C values that
a parabola with those 3 points would have had.
I made a method that takes in an X value and returns a Vector3 containing the point this X is on a linear equation, but instead, changing the vector's Y value based on the parabola's equation.
Practically creating an elevated line, I made something that looks and behaves perfectly like a 3D parabola.
If you're in C#, here is the code(images):
FIX!!
in the Linear's GetX(float x) method.
it should be:
public Vector3 GetX(float x) => => r0 + (x - r0.x)/v.x * v;
I made a slight mistake in the calculations which I noticed immediately and changed.
I have two positions and want to calculate the position where the red star is. (I want to move B to the "red star location" but I don't know the coordinates.)
I have the position of A and B and a minimum distance from position A. So my question is how do I calculate the nearest position to B within the specified distance.
Hope someone understand what i'm trying to accomplish.
Assuming that:
You want B to be in the same direction from A as before
You want to move B to a specific distance from A
If B is farther away, move it closer to A to get to the distance
If B is closer than the specific distance, move it away from A to get to the distance
Then this is the way to do it:
Calculate the current distance from A to B
Calculate the difference in position between A and B (in 2 or 3 dimensions)
Divide this difference by a ratio that is "current distance / wanted distance"
For instance, if the current distance is 2x as far away from A as you want, you would divide the difference in position by 2
Set the new position of B to be A + the new differences
Here's some sample code:
var wantedDistance = 40.0;
var distance = Math.Sqrt((A.x - B.x)*(A.x - B.x) + (A.y - B.y)*(A.y - B.y));
var diffX = B.x - A.x;
var diffY = B.y - A.y;
var ratio = distance / wantedDistance;
var newDiffX = diffX / ratio;
var newDiffY = diffY / ratio;
var newB = new PointF(A.x + newDiffX, A.y + newDiffY);
In a comment you say that if B is inside the radius, then it is in range so it should not be moved. You would simply handle this by comparing distance with wantedDistance and if lower, then you're done. Simply add this line of code after the var distance = ... line:
if (distance < wantedDistance)
return B; // or whatever you want to do when "done"
If you need to do this in 3D space, simply augment every calculation and diff to handle the Z dimension as well.
I am an architecture student trying to solve a spatial problem with C# in Grasshopper for Rhino.
The space I am trying to create is an exhibition space in an airport. The space will be made up of elements of similar length. The idea is to connect them with a hinge and thereby allow them to create spaces of different layout and size according to how many elements are used.
As you can see from the illustration I would like the space to end with an opening an element length away from the starting point.
My first attempt has been to create equilateral triangles depending on the number of segments (walls) needed.
In short, from the starting point, triangles are created, and then the sides of the triangle that form the outer border are added to a list of points. This point list is returned to the Grasshopper application, which draws lines between the points. A little point is that I made the creation of the next triangle randomly either from the side AC or BC from the last triangle.
Here is an example of the spaces created (for 12 - 8 - 14 - 20 elements):
Here is the source code that creates these point lists:
private void RunScript(double radius, int walls, ref object A)
{
//
List<Point3d> pointList = new List<Point3d>();
List<Point3d> lastList = new List<Point3d>();
bool alternate = true;
bool swapped = false;
Random turn = new Random();
// set up the first part of the triangle
Point3d point1 = new Point3d(0, 0, 0);
Point3d point2 = new Point3d(0, radius, 0);
pointList.Add(point1);
pointList.Add(point2);
Point3d calcPoint;
for(int i = 0; i < walls - 1; i++) // walls - 1, is because I need one less triangle to get to the amount of walls
{
// use the method to create two similar circles and return the intersection point
// in order to create an equilateral triangle
calcPoint = FindCircleIntersections(point1.X, point1.Y, point2.X, point2.Y, radius, alternate);
// random generator: will decide if the new triangle should be created from side BC or AC
bool rotate = turn.Next(2) != 0;
Print("\n" + rotate);
// set the 2nd and 3rd point as 1st and 2nd - depending on random generator.
if(rotate)
{
point1 = point2;
if(swapped == true)
swapped = false;
else
swapped = true;
}
// if the direction is swapped, the next point created will not be a part of the outer border
if(swapped)
lastList.Add(calcPoint);
else
pointList.Add(calcPoint);
point2 = calcPoint;
// swap direction of intersection
if(rotate)
{
if(alternate)
alternate = false;
else
alternate = true;
}
}
lastList.Reverse();
foreach (Point3d value in lastList)
{
pointList.Add(value);
}
A = pointList;
}
// Find the points where the two circles intersect.
private Point3d FindCircleIntersections(
double cx0, double cy0, double cx1, double cy1, double rad, bool alternate)
{
// Find the distance between the centers.
double dx = cx0 - cx1;
double dy = cy0 - cy1;
double dist = Math.Sqrt(dx * dx + dy * dy);
// Find a and h.
double a = (rad * rad - rad * rad + dist * dist) / (2 * dist);
double h = Math.Sqrt(rad * rad - a * a);
// Find P2.
double cx2 = cx0 + a * (cx1 - cx0) / dist;
double cy2 = cy0 + a * (cy1 - cy0) / dist;
// Get the points P3.
if(alternate)
return new Point3d((double) (cx2 + h * (cy1 - cy0) / dist), (double) (cy2 - h * (cx1 - cx0) / dist), 0);
else
return new Point3d((double) (cx2 - h * (cy1 - cy0) / dist), (double) (cy2 + h * (cx1 - cx0) / dist), 0);
}
What I would like to do, is to vary the creation of these shapes, so that they are not only corridors, but resemble my initial sketches. I would like an algorithm to take an input of segments (number and length) and then propose different space layouts which are possible to create with this number of segments. I guess because of tesselation the space would have to be created with triangles, squares or hexagons? Do you think I should look into this "maximum area" algorithm : Covering an arbitrary area with circles of equal radius here on stackoverflow?
I would greatly appreciate any help on this algorithm. Cheers, Eirik
If you're merely interested in a program to generate instances to be externally evaluated (and not all such instances), you could "inflate" your curve. For example, in the 14-segment instance in your second image, there is a place where the curve goes inward and doubles back -- so your list of points has one point repeated. For curves like this you could cut out everything between the two (identical) points (A and B), as well as one of the surrounding points (A or B), and you have reclaimed some points to expand your curve - possibly resulting in a non-corridor structure. You may have to work some magic to ensure it is a "closed" curve, though, buy alternately adding segments to the front and the back of the curve.
Another opportunity: if you can identify the curve's interior (and there are algorithms for this), then anywhere that two segments form a concave angle with respect to your curve, you could blow it out to make a non-corridorish area. E.g. the second and third segments of your 14-segment curve above could be blown out to the left.
Successively applying these two methods to your corridor-like curve should generate many of the shapes you're looking for. Good luck!