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.
Related
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 a problem and please let me know if my solution is correct.
I have a known point, at location A(x1,y1,z1) and the origin O(0,0,0) and I would like to find the coordinates of a point B(x2,y2,z2) that is located on the line OA, and the distance OB is 1.2 times greater then OA.
So, my idea is to obtain the equation of the line formed by points O and A.
The direction of OA is (-x1, -y1, -z1), so the equation of the line is:
x = -x1*t;
y = -y1*t;
z = -z1*t;
Distance OA is sqrt( (x1-0)^2 + (y1-0)^2 + (z1-0)^2). KNOWN
Distance OB is sqrt( (x2-0)^2 + (y2-0)^2 + (z2-0)^2). UNKNOWN
I can replace the x, y, z points determined for the line equation in the distance OB, and the result should be 1.2 times greater then the distance OA.
So, sqrt( (-x1*t-0)^2 + (-y1*t-0)^2 + (-z1*t-0)^2) = 1.2 * dist(OA).
I find t from here, solving the quadratic equation and I obtain the coordinates of the point by replacing the t in the equation of the line.
Is this correct?
Thank you for your time.
EDIT:
This is my code:
rangeRatio = 1.114;
norm = sqrt((P2(1) - P1(1))^2 + (P2(2) - P1(2))^2 + (P2(3) - P1(3))^2);
P3(1) = P1(1) + ((P2(1,1) - P1(1)) /norm) * rangeRatio;
P3(2) = P1(2) + ((P2(1,2) - P1(2)) /norm) * rangeRatio;
P3(3) = P1(3) + ((P2(1,3) - P1(3)) /norm) * rangeRatio;
I tried also norm = 1, and i get slightly different results but still not always colinear.
Thank you
It is even a lot easier; you can just multiply a, b and c by 1.2. This gives a line that is 1.2 times the size of the original line.
Ok so let me try to explain this the best way that i can.
I have two points plotted 'A' and 'B' and I am trying to plot a third point 'C' so that it is past point 'B' but along the same slope. I have the angle of the line and I would post some code but I really have no idea where to begin.
any help would be awesome!
Just a little code that i do have
CGPoint vector = ccpSub(touchedPoint, fixedPoint);
CGFloat rotateAngle = -ccpToAngle(vector);
Assuming that by this you mean you need a 3rd point C added such that all the points are colinear, all you need to do is calculate the vector that takes you from A to B, and then generate a new point by adding multiples of this vector to the point B. Choose the multiple based on the distance you want C to be from B.
As an example, say A = (2,2), B = (4,3). Then the vector from A to B is given by (2,1).
All you need to do then is work out how far your new point is from B and add a multiple K*(2,1) to your point B where K is chosen to meet the requirements of your distance
I am assuming you are in 2D, but the same method would apply in higher dimensions
My math is rusty, but the linear equation is generally represented as y=m*x+b, where m is the slope, and b is the y-intercept. You can get m, the slope, by taking the difference of the y values and dividing that by the difference in the x values, e.g., if A = (2,2) and B = (4,3), then m is (3-2)/(4-2) or 0.5. Then, you can solve the linear equation for b, the y-intercept, i.e. b=y-m*x and then plug in either of the data points, e.g. if we plug in the x and y values for point A, you get b = 2 - 0.5 * 2 = 1. Now knowing the slope, m (0.5 in this example), and the y-intercept, b (1 in this example), you can calculate the y for any x value using y=m*x+b, in this case y=0.5*x+1.
So, if touchedPoint and fixedPoint are CGPoint, you can calculate the slope and y-intercept from fixedPoint and touchedPoint like so:
double m = (fixedPoint.y - touchedPoint.y) / (fixedPoint.x - touchedPoint.x);
double b = fixedPoint.y - m * fixedPoint.x;
Now, you don't say how you want to determine where this third point, C, is. But if you, for example, knew the x coordinate for this new point C, you can calculate the y coordinate that falls on the same line as follows:
CGPoint pointC;
pointC.x = 400; // or set this to whatever you want
pointC.y = m * pointC.x + b;
I am processing a series of points which all have the same Y value, but different X values. I go through the points by incrementing X by one. For example, I might have Y = 50 and X is the integers from -30 to 30. Part of my algorithm involves finding the distance to the origin from each point and then doing further processing.
After profiling, I've found that the sqrt call in the distance calculation is taking a significant amount of my time. Is there an iterative way to calculate the distance?
In other words:
I want to efficiently calculate: r[n] = sqrt(x[n]*x[n] + y*y)). I can save information from the previous iteration. Each iteration changes by incrementing x, so x[n] = x[n-1] + 1. I can not use sqrt or trig functions because they are too slow except at the beginning of each scanline.
I can use approximations as long as they are good enough (less than 0.l% error) and the errors introduced are smooth (I can't bin to a pre-calculated table of approximations).
Additional information:
x and y are always integers between -150 and 150
I'm going to try a couple ideas out tomorrow and mark the best answer based on which is fastest.
Results
I did some timings
Distance formula: 16 ms / iteration
Pete's interperlating solution: 8 ms / iteration
wrang-wrang pre-calculation solution: 8ms / iteration
I was hoping the test would decide between the two, because I like both answers. I'm going to go with Pete's because it uses less memory.
Just to get a feel for it, for your range y = 50, x = 0 gives r = 50 and y = 50, x = +/- 30 gives r ~= 58.3. You want an approximation good for +/- 0.1%, or +/- 0.05 absolute. That's a lot lower accuracy than most library sqrts do.
Two approximate approaches - you calculate r based on interpolating from the previous value, or use a few terms of a suitable series.
Interpolating from previous r
r = ( x2 + y2 ) 1/2
dr/dx = 1/2 . 2x . ( x2 + y2 ) -1/2 = x/r
double r = 50;
for ( int x = 0; x <= 30; ++x ) {
double r_true = Math.sqrt ( 50*50 + x*x );
System.out.printf ( "x: %d r_true: %f r_approx: %f error: %f%%\n", x, r, r_true, 100 * Math.abs ( r_true - r ) / r );
r = r + ( x + 0.5 ) / r;
}
Gives:
x: 0 r_true: 50.000000 r_approx: 50.000000 error: 0.000000%
x: 1 r_true: 50.010000 r_approx: 50.009999 error: 0.000002%
....
x: 29 r_true: 57.825065 r_approx: 57.801384 error: 0.040953%
x: 30 r_true: 58.335225 r_approx: 58.309519 error: 0.044065%
which seems to meet the requirement of 0.1% error, so I didn't bother coding the next one, as it would require quite a bit more calculation steps.
Truncated Series
The taylor series for sqrt ( 1 + x ) for x near zero is
sqrt ( 1 + x ) = 1 + 1/2 x - 1/8 x2 ... + ( - 1 / 2 )n+1 xn
Using r = y sqrt ( 1 + (x/y)2 ) then you're looking for a term t = ( - 1 / 2 )n+1 0.36n with magnitude less that a 0.001, log ( 0.002 ) > n log ( 0.18 ) or n > 3.6, so taking terms to x^4 should be Ok.
Y=10000
Y2=Y*Y
for x=0..Y2 do
D[x]=sqrt(Y2+x*x)
norm(x,y)=
if (y==0) x
else if (x>y) norm(y,x)
else {
s=Y/y
D[round(x*s)]/s
}
If your coordinates are smooth, then the idea can be extended with linear interpolation. For more precision, increase Y.
The idea is that s*(x,y) is on the line y=Y, which you've precomputed distances for. Get the distance, then divide it by s.
I assume you really do need the distance and not its square.
You may also be able to find a general sqrt implementation that sacrifices some accuracy for speed, but I have a hard time imagining that beating what the FPU can do.
By linear interpolation, I mean to change D[round(x)] to:
f=floor(x)
a=x-f
D[f]*(1-a)+D[f+1]*a
This doesn't really answer your question, but may help...
The first questions I would ask would be:
"do I need the sqrt at all?".
"If not, how can I reduce the number of sqrts?"
then yours: "Can I replace the remaining sqrts with a clever calculation?"
So I'd start with:
Do you need the exact radius, or would radius-squared be acceptable? There are fast approximatiosn to sqrt, but probably not accurate enough for your spec.
Can you process the image using mirrored quadrants or eighths? By processing all pixels at the same radius value in a batch, you can reduce the number of calculations by 8x.
Can you precalculate the radius values? You only need a table that is a quarter (or possibly an eighth) of the size of the image you are processing, and the table would only need to be precalculated once and then re-used for many runs of the algorithm.
So clever maths may not be the fastest solution.
Well there's always trying optimize your sqrt, the fastest one I've seen is the old carmack quake 3 sqrt:
http://betterexplained.com/articles/understanding-quakes-fast-inverse-square-root/
That said, since sqrt is non-linear, you're not going to be able to do simple linear interpolation along your line to get your result. The best idea is to use a table lookup since that will give you blazing fast access to the data. And, since you appear to be iterating by whole integers, a table lookup should be exceedingly accurate.
Well, you can mirror around x=0 to start with (you need only compute n>=0, and the dupe those results to corresponding n<0). After that, I'd take a look at using the derivative on sqrt(a^2+b^2) (or the corresponding sin) to take advantage of the constant dx.
If that's not accurate enough, may I point out that this is a pretty good job for SIMD, which will provide you with a reciprocal square root op on both SSE and VMX (and shader model 2).
This is sort of related to a HAKMEM item:
ITEM 149 (Minsky): CIRCLE ALGORITHM
Here is an elegant way to draw almost
circles on a point-plotting display:
NEW X = OLD X - epsilon * OLD Y
NEW Y = OLD Y + epsilon * NEW(!) X
This makes a very round ellipse
centered at the origin with its size
determined by the initial point.
epsilon determines the angular
velocity of the circulating point, and
slightly affects the eccentricity. If
epsilon is a power of 2, then we don't
even need multiplication, let alone
square roots, sines, and cosines! The
"circle" will be perfectly stable
because the points soon become
periodic.
The circle algorithm was invented by
mistake when I tried to save one
register in a display hack! Ben Gurley
had an amazing display hack using only
about six or seven instructions, and
it was a great wonder. But it was
basically line-oriented. It occurred
to me that it would be exciting to
have curves, and I was trying to get a
curve display hack with minimal
instructions.