Given a convex 3d polygon (convex hull) How can I determine the correct direction for normal surface/vertex vectors? As the polygon is convex, by correct I mean outward facing (away from the centroid).
def surface_normal(centroid, p1, p2, p3):
a = p2-p1
b = p3-p1
n = np.cross(a,b)
if **test including centroid?** :
return n
else:
return -n # change direction
I actually need the normal vertex vectors as I am exporting as a .obj file, but I am assuming that I would need to calculate the surface vectors before hand and combine them.
This solution should work under the assumption of a convex hull in 3d. You calculate the normal as shown in the question. You can normalize the normal vector with
n /= np.linalg.norm(n) # which should be sqrt(n[0]**2 + n[1]**2 + n[2]**2)
You can then calculate the center point of your input triangle:
pmid = (p1 + p2 + p3) / 3
After that you calculate the distance of the triangle-center to your surface centroid. This is
dist_centroid = np.linalg.norm(pmid - centroid)
The you can calculate the distance of your triangle_center + your normal with the length of the distance to the centroid.
dist_with_normal = np.linalg.norm(pmid + n * dist_centroid - centroid)
If this distance is larger than dist_centroid, then your normal is facing outwards. If it is smaller, it is pointing inwards. If you have a perfect sphere and point towards the centroid, it should almost be zero. This may not be the case for your general surface, but the convexity of the surface should make sure, that this is enough to check for its direction.
if(dist_centroid < dist_with_normal):
n *= -1
Another, nicer option is to use a scalar product.
pmid = (p1 + p2 + p3) / 3
if(np.dot(pmid - centroid, n) < 0):
n *= -1
This checks if your normal and the vector from the mid of your triangle to the centroid have the same direction. If that is not so, change the direction.
Related
What I have tried already: d = |v||PQ|sin("Theta")
Now, I need to determine what theta is, so I set up a position on a makeshift graph, the graph I made was on the xy plane only as the z plane complicates things needlessly for finding theta. So, I ended up with an acute angle, and if the angle is acute, then I have to find theta which according to dot product facts is greater than 0.
I do not have access to theta, so I used the same princples from cross dots. u * v = |u||v|cos("theta") but in this case, u and v are PQ and v. A vector is a vector, right?
so now I have theta = acos((v*PQ)/(|v||PQ))
with that I get (4sqrt(10))/15 = 32.5125173162 in degrees, so the angle is 32.5125173162 degrees.
So, now that I have theta, I plug it into my distance formula |v||PQ|sin(32.5125173162)
3*sqrt(10)*sin(32.5125173162) = 5.0990195136
or for the sake of simplicity, 5.1
I however want to know if this question is correct.
If it is NOT correct, what can I do to correct it? At what points did I use incorrect information?
This is not a question with a definitive answer in the back of the book, its a question on the side of a page that said: "try this!"
There are a couple of problems with this question.
From the context it looks like you mean for both v and PQ to be vectors. The "distance" between two vectors is an awkward (not well defined) question because vectors are not position bound.
You are using the cross product formula and I have no idea why:
|AxB| = |A||B|Sin(theta)
I think what you are actually trying to do is calculate the distance between the terminal points of the vectors, (2, 1, 2) and (1, 0, 3). Just use the Pythagorean Theorem (extended to 3D) for this.
d = sqrt( (x1 - x2)^2 + (y1 - y2)^2 + (z1 - z2)^2 )
d = sqrt( (2 - 1)^2 + (1 - 2)^2 + (2 - 3)^2 )
d = sqrt( 1^2 + (-1)^2 + (-1)^2 )
d = sqrt(3)
Edit:
If what you need really is the magnitude of the cross product, |AxB| then just find the cross product (using the determinant) and then calculate the magnitude of the result. There is no need for the formula you were using.
I'm a little confused about the difference between the DLT algorithm described here and the homography estimation described here. In both of these techniques, we are trying to solve for the entries of a 3x3 matrix by using at least 4 point correspondences. In both methods, we set up a system where we have a "measurement" matrix and we use SVD to solve for the vector of elements that make up H. I was wondering why there are two techniques that seem to do the same thing, and why one might be used over the other.
You have left-right image correspondences {p_i} <-> {p'_i}, where p_i = (x_i, y_i), etc.
Normalizing them to the unit square means computing two shifts m=(mx, my), m'=(mx', my'), and two scales s=(sx,sy), s'=(sx',sy') such that q_i = (p_i - m) / s and q_i' = (p_i' - m') / s', and both the {q_i} and the {q'_i} transformed image points are centered at (0,0) and approximately contained within a square of unit side length. A little math shows that a good choice for the m terms are the averages of the x,y coordinates in each set of image points, and for the s terms you use the standard deviations (or twice the standard deviation) times 1/sqrt(2).
You can express this normalizing transformation in matrix form: q = T p,
where T = [[1/sx, 0, -mx/sx], [0, 1/sy, -my/sy], [0, 0, 1]], and likewise q' = T' p'.
You then compute the homography K between the {q_i} and {q'_i} points: q_i' = K q_i.
Finally, you denormalize K into the origina (un-normalized) coordinates thus: H = inv(T') K T, and H is the desired homography that maps {p} into {p'}.
I have a straight line in space with an start and end point (x,y,z) and I am attempting to get the angle between this vector and the plane defined by z=0. I am using VB.NET
Here is a picture of the line in my 3d environment (the line I'm intersted in is circled in red) :
It is set to an angle of 70 degrees right now.
You need 2 rays to define an angle.
If you want the angle between a vector and a plane, it is defined for any vector in that plane. However, there is only one minimal value for that, which is the angle between a vector and its projection onto said plane.
Therefore, that minimal value is the one we take when we speak of the angle between a vector and a plane.
This value is also π/2 - the angle between your vector and the the vector that is normal to the plane.You can read more about it all on this site.
With v your vector (thus v.x = end.x - start.x and idem for y and z), n the normal to the plane and a the angle you are looking for, we know from the definition of a scalar product that:
<v,n> = ||v|| * ||n|| * cos(π/2 - a)
We know cos(π/2 - a) = sin(a), and the normal to the z=0 plane is simply the vector n = (0, 0, 1). Thus both the scalar product, v.x * n.x + v.y * n.y + v.z * n.z, and the norm of n, ||n|| = 1, can be simplified a lot. We get the following expression:
sin(a) = v.z / ||v||
Thus finally, the formula by taking the reciprocical of the sine and expliciting the norm of v:
a = Asin(v.z / sqrt( v.x*v.x + v.y*v.y + v.z*v.z ))
According to this documentation the Asin function exists in your System.Math class. It does, however, return the value in radians:
Return Value
Type: System.Double
An angle, θ, measured in radians, such that -π/2 ≤ θ ≤ π/2
-or-
NaN if d < -1 or d > 1 or d equals NaN.
Luckily the same System.Math class contains the value of π so that you can do the conversion:
a *= 180 / Math.PI
I have solid object that is spinning with a torque W, and I want to calculate the force F applied on a certain point that's D units away from the center of the object. All these values are represented in Vector3 format (x, y, z)
I know until now that W = D x F, where x is the cross product, so by expanding this I get:
Wx = Dy*Fz - Dz*Fy
Wy = Dz*Fx - Dx*Fz
Wz = Dx*Fy - Dy*Fx
So I have this equation, and I need to find (Fx, Fy, Fz), and I'm thinking of using the Simplex method to solve it.
Since the F vector can also have negative values, I split each F variable into 2 (F = G-H), so the new equation looks like this:
Wx = Dy*Gz - Dy*Hz - Dz*Gy + Dz*Hy
Wy = Dz*Gx - Dz*Hx - Dx*Gz + Dx*Hz
Wz = Dx*Gy - Dx*Hy - Dy*Gx + Dy*Hx
Next, I define the simplex table (we need <= inequalities, so I duplicate each equation and multiply it by -1.
Also, I define the objective function as: minimize (Gx - Hx + Gy - Hy + Gz - Hz).
The table looks like this:
Gx Hx Gy Hy Gz Hz <= RHS
============================================================
0 0 -Dz Dz Dy -Dy <= Wx = Gx
0 0 Dz -Dz -Dy Dy <= -Wx = Hx
Dz -Dz 0 0 Dx -Dx <= Wy = Gy
-Dz Dz 0 0 -Dx Dx <= -Wy = Hy
-Dy Dy Dx -Dx 0 0 <= Wz = Gz
Dy -Dy -Dx Dx 0 0 <= -Wz = Hz
============================================================
1 -1 1 -1 1 -1 0 = Z
The problem is that when I run it through an online solver I get Unbounded solution.
Can anyone please point me to what I'm doing wrong ?
Thanks in advance.
edit: I'm sure I messed up some signs somewhere (for example the Z should be defined as a max), but I'm sure I'm wrong when defining something more important.
There exists no unique solution to the problem as posed. You can only solve for the tangential projection of the force. This comes from the properties of the vector (cross) product - it is zero for collinear vectors and in particular for the vector product of a vector by itself. Therefore, if F is a solution of W = r x F, then F' = F + kr is also a solution for any k:
r x F' = r x (F + kr) = r x F + k (r x r) = r x F
since the r x r term is zero by the definition of vector product. Therefore, there is not a single solution but rather a whole linear space of vectors that are solutions.
If you restrict the solution to forces that have zero projection in the direction of r, then you could simply take the vector product of W and r:
W x r = (r x F) x r = -[r x (r x F)] = -[(r . F)r - (r . r)F] = |r|2F
with the first term of the expansion being zero because the projection of F onto r is zero (the dot denotes scalar (inner) product). Therefore:
F = (W x r) / |r|2
If you are also given the magnitude of F, i.e. |F|, then you can compute the radial component (if any) but there are still two possible solutions with radial components in opposing directions.
Quick dirty derivation...
Given D and F, you get W perpendicular to them. That's what a cross product does.
But you have W and D and need to find F. This is a bad assumption, but let's assume F was perpendicular to D. Call it Fp, since it's not necessarily the same as F. Ignoring magnitudes, WxD should give you the direction of Fp.
This ignoring magnitudes, so fix that with a little arithmetic. Starting with W=DxF applied to Fp:
mag(W) = mag(D)*mag(Fp) (ignoring geometry; using Fp perp to D)
mag(Fp) = mag(W)/mag(D)
Combining the cross product bit for direction with this stuff for magnitude,
Fp = WxD / mag(WxD) * mag(Fp)
Fp = WxD /mag(W) /mag(D) *mag(W) /mag(D)
= WxD / mag(D)^2.
Note that given any solution Fp to W=DxF, you can add any vector proportional to D to Fp to obtain another solution F. That is a totally free parameter to choose as you like.
Note also that if the torque applies to some sort of axle or object constrained to rotate about some axis, and F is applied to some oddball lever sticking out at a funny angle, then vector D points in some funny direction. You want to replace D with just the part perpendicular to the axle/axis, otherwise the "/mag(D)" part will be wrong.
So from your comment is clear that all rotations are spinning around center of gravity
in that case
F=M/r
F force [N]
M torque [N/m]
r scalar distance between center of rotation [m]
this way you know the scalar size of your Force
now you need the direction
it is perpendicular to rotation axis
and it is the tangent of the rotation in that point
dir=r x axis
F = F * dir / |dir|
bolds are vectors rest is scalar
x is cross product
dir is force direction
axis is rotation axis direction
now just change the direction according to rotation direction (signum of actual omega)
also depending on your coordinate system setup
so ether negate F or not
but this is in 3D free rotation very unprobable scenario
the object had to by symmetrical from mass point of view
or initial driving forces was applied in manner to achieve this
also beware that after first hit with any interaction Force this will not be true !!!
so if you want just to compute Force it generate on certain point if collision occurs is this fine
but immediately after this your spinning will change
and for non symmetric objects the spinning will be most likely off the center of gravity !!!
if your object will be disintegrated then you do not need to worry
if not then you have to apply rotation and movement dynamics
Rotation Dynamics
M=alpha*I
M torque [N/m]
alpha angular acceleration
I quadratic mass inertia for actual rotation axis [kg.m^2]
epislon''=omega'=alpha
' means derivation by time
omega angular speed
epsilon angle
I have successfully drawn a Quad2D or Bezier curve in java. I have the equation for the same. But I need to determine whether a particular point(x,y) lies on the curve or not. I tried using Quad2D.contains and a few GeneralPath APIs, I could not get the result accurately.
Can someone help to find out the solution to this?
I think you've meant QuadCurve2D class, which is quadratic Bezier curve.
There seems to be no ready-made method for that, and the problem comes to the distance from the point to the Bezier curve.
Let's P0 will be your point, P1 is a start point, P2 - a control point and P3 is an end point of your curve.
Then point on the curve will be given by
P = B(t)
There is such t, for which distance between P and P0 will be minimal.
F(t) = (B(t)_x - P0_x)^2 + (B(t)_y - P0_y)^2 -> min
If distance is 0 or less than certain error, then P0 is on the curve.
t can be found with Netwon's iterative method by minimizing cost function
t_n = t_n-1 + F'(t) / F''(t)
where F' is first derivative of cost function and F'' is its second derivative.
F'(t) = 2 * (B(t)_x - P0_x) * B'(t)_x + 2 * (B(t)_y - P0_y) * B'(t)_y
F''(t) = 2 * B'(t)_x * B'(t)_x + 2 * (B(t)_x - P0_x) * B''(t)_x +
2 * B'(t)_y * B'(t)_y + 2 * (B(t)_y - P0_x) * B''(t)_y
First derivative of quadratic Bezier curve B'(t) is a line segment with a start point (P2 - P1) and end point (P3 - P2). Second derivative B''(t) is a point P3 - 2 * P2 + P1.
Plugging everything together will give a formula to t for which F(t) is minimal.