I watched and tried to understand bunch of sites and videos about this, and I came into a weird conclusions and some questions. I need some help to explain which one of the method is right, or even both of them are right (but I got different result from each methods).
I'm sorry that I'm bad at explaining things, the first method is solve the equations normally. But, here the link for the video I tried to learn from
https://www.youtube.com/watch?v=o7CfCDkRwfY
Second method is to do cross product for the direction and find the point by set one of the variables as 0. https://www.youtube.com/watch?v=jozabh0lFmo
I tried for this example
x+2y+z−1=0
2x+3y−2z+2=0
and turned out for different answers. Is both of the method are correct, or which one? Thank you.
You have two equations with three unknowns.
You can eliminate one variable and solve for a relationship between the remaining two. Let's eliminate z.
Multiply the first equation by 2:
2x + 4y + 2z = 2
Add this to the second equation:
4x + 7y = 0
You can solve for y as a function of x:
y = -4x/7
Substitute this back into the first equation:
x - 8x/7 + z = 1
Simplify by combining the first and second terms:
-x/7 + z = 1
Solve for z:
z = 1 + x/7
Now you have an equation for the line in 3D space.
-inf <= x <= +inf
y = -4x/7
z = 1 + x/7
Both your equations are satisfied by these two points. Since two points are enough to define a line in Euclidean space I'd say I've got the correct answer.
This line goes through the point (0, 0, 1). It also goes through (7, -4, 2)
Here's a parametric representation of that line for -inf <= t <= +inf:
(x, y, z) = (0, 0, 1) + t*(7, -4, 1)
You can see that when t = 0 (x, y, z) = (0, 0, 1) (first point above) and when t = 1 (x, y, z) = (7, -4, 2) (second point above).
I didn't look at either of your videos. This is how I'd solve it.
This is high school algebra.
Related
I have two points which form one line: (1,4) and (3,6), and another two which form another line: (2,1) and (4,2). These lines are continuous and I can find their intersection points by finding the equation for each line, and then equating them to find the x value at the intersection point, and then the y value.
i.e. for the first line, the equation is y = x + 3, and the second is y = 0.5x. At the intersection the y values are the same so x + 3 = 0.5x. So x = -6. Subbing this back into either of the equations gives a y value of -3.
From those steps, I now know that the intersection point is (-6,-3). The problem is I need to do the same steps in Excel, preferably as one formula. Can anyone give me some advice on how I would start this?
Its long but here it is:
Define x1,y1 and x2,y2 for the 1st line and x3,y3 and x4,y4 for the second.
x = (x2y1-x1y2)(x4-x3)-(x4y3-x3y4)(x2-x1) / [ (x2-x1)(y4-y3) - (x4-x3)(y2-y1) ]
y = (x2y1-x1y2)(y4-y3)-(x4y3-x3y4)(y2-y1) / [ (x2-x1)(y4-y3) - (x4-x3)(y2-y1) ]
Note that the denominators are the same. They will be ZERO! when the system has no solution. So you may want to check that in another cell and conditionally compute the answer.
Essentially, this formula is derived by solving a system of equations for x and y by hand using generic points (x1,y1), (x2,y2), (x3,y3), and (x4,y4). Easier yet, is solving the system by hand using well developed linear algebra concepts.
Wikipedia outlines this procedure well: Line-line intersection.
Also, this website describes all the different formulas and lets you put in whatever data you have in any mixed format and provides many details of the solutions: Everything about 2 lines.
Here's a matrix based solution:
x - y = -3
0.5*x - y = 0
Written as a matrix equation (I apologize for the poor typesetting):
| 1.0 -1.0 |{ x } { -3 }
| 0.5 -1.0 |{ y } = { 0 }
You can invert this matrix or use LU decomposition to solve it to get the answer. That method will work for any number of cases where you have one equation for each unknown.
This is easy to do by hand:
Subtract the second equation from the first: 0.5*x = -3
Divide both sides by 0.5: x = -6
Substitute this result into the other equation: y = 0.5*x = -3
I am using this octave code for solving differential equation.
# Define the right-hand side of the equation:
xvall= -11 ;#xvall
xvalu= 10 ;#xvalu
range=5000;
function ret=f(x,t);ret= t ;end;
# ywill be the values of the function at these moments of time.
t=linspace(xvall,xvalu,range);
y=lsode ('f', 2, linspace(xvall,xvalu,range));
y
plot(t,y);
i got the graph like this .
But when the same conditions are passed to wolfram alpha
I am getting the graph from 60 to 0 for y value
graph is
why is the graph behaving differently in two situations.
https://www.wolframalpha.com/input/?i=Runge-Kutta+method%2C+dy%2Fdx+%3D+x%2C+y%280%29+%3D+2%2C+from+-11+to+10%2C+h+%3D+0.25
To specify an initial value problem for an ordinary differential equation you need to define the initial condition. Here for Octave you have specified x(-11) = 2 since xvall = -11 and for Wolfram Alpha you have specified y(0) = 2. That is why you have two different solutions.
Octave
Octave's lsode (f,x_0,ts) solves the following initial value problem
dx/dt = t
x(t_0) = x_0
t in ts
Here ts is specified as a set of points in the interval [t_0,t_1]. You have specified t_0 = -11, t_1 = 10.
In closed form the solution to this problem is x = (t^2 - 117) / 2
Wolfram
For Wolfram you have used the semi-formal syntax:
Runge-Kutta method, dy/dx = x, y(0) = 2, from -11 to 10, h = 0.25
In closed form the solution to this problem would be y = (x^2 + 4) /2
The corresponding initial value problem is clearly different. Hence different results.
I'm trying to generate some axis vectors from parameters commonly used to specify crystallographic unit cells. These parameters consist of the length of the three axes: a,b,c and the angles between them: alpha,beta,gamma. By convention alpha is the angle between the b and c axes, beta is between a and c, and gamma between a and b.
Now getting vector representations for the first two is easy. I can arbitrarily set the the a axis to the x axis, so a_axis = [a,0,0]. I then need to rotate b away from a by the angle gamma, so I can stay in the x-y plane to do so, and b_axis = [b*cos(gamma),b*sin(gamma),0].
The problem is the third vector. I can't figure out a nice clean way to determine it. I've figured out some different interpretations but none of them have panned out. One is imagining the there are two cones around the axes axis_a and axis_b whose sizes are specified by the angles alpha and beta. The intersection of these cones create two lines, the one in the positive z direction can be used as the direction for axis_c, of length c.
Does someone know how I should go about determining the axis_c?
Thanks.
The angle alpha between two vectors u,v of known length can be found from their inner (dot) product <u,v>:
cos(alpha) = <u,v>/(||u|| ||v||)
That is, the cosine of alpha is the inner product of the two vectors divided by the product of their lengths.
So the z-component of your third can be any nonzero value. Scaling any or all of the axis vectors after you get the angles right won't change the angles, so let's assume (say) Cz = 1.
Now the first two vectors might as well be A = (1,0,0) and B = (cos(gamma),sin(gamma),0). Both of these have length 1, so the two conditions to satisfy with choosing C are:
cos(alpha) = <B,C>/||C||
cos(beta) = <A,C>/||C||
Now we have only two unknowns, Cx and Cy, to solve for. To keep things simple I'm going to just refer to them as x and y, i.e. C = (x,y,1). Thus:
cos(alpha) = [cos(gamma)*x + sin(gamma)*y]/sqrt(x^2 + y^2 + 1)
cos(beta) = x/(sqrt(x^2 + y^2 + 1)
Dividing the first equation by the second (assuming beta not a right angle!), we get:
cos(alpha)/cos(beta) = cos(gamma) + sin(gamma)*(y/x)
which is a linear equation to solve for the ratio r = y/x. Once you have that, substituting y = rx in the second equation above and squaring gives a quadratic equation for x:
cos^2(beta)*((1+r^2)x^2 + 1) = x^2
cos^2(beta) = (1 - cos^2(beta)*(1 + r^2))x^2
x^2 = cos^2(beta)/[(1 - cos^2(beta)*(1 + r^2))]
By squaring the equation we introduced an artifact root, corresponding to choosing the sign of x. So check the solutions for x you get from this in the "original" second equation to make sure you get the right sign for cos(beta).
Added:
If beta is a right angle, things are simpler than the above. x = 0 is forced, and we have only to solve the first equation for y:
cos(alpha) = sin(gamma)*y/sqrt(y^2 + 1)
Squaring and multiplying away the denominator gives a quadratic for y, similar to what we did before. Remember to check your choice of a sign for y:
cos^2(alpha)*(y^2 + 1) = sin^2(gamma)*y^2
cos^2(alpha) = [sin^2(gamma) - cos^2(alpha)]*y^2
y^2 = cos^2(alpha)/[sin^2(gamma) - cos^2(alpha)]
Actually if one of the angles alpha, beta, gamma is a right angle, it might be best to label that angle gamma (between the first two vectors A,B) to simplify the computation.
Here is a way to find all Cx, Cy, Cz (first two are the same as in the other answer), given that A = (Ax,0,0), B = (Bx, By, 0), and assuming that |C| = 1
1) cos(beta) = AC/(|A||C|) = AxCx/|A| => Cx = |A|cos(beta)/Ax = cos(beta)
2) cos(alpha) = BC/(|B||C|) = (BxCx+ByCy)/|B| => Cy = (|B|cos(alpha)-Bx cos(beta))/By
3) To find Cz let O be the point at (0,0,0), T the point at (Cx,Cy,Cz), P be the projection of T on Oxy and Q be the projection of T on Ox. So P is the point at (Cx,Cy,0) and Q is the point at (Cx,0,0). Thus from the right angle triangle OQT we get
tan(beta) = |QT|/||OQ| = |QT|/Cx
and from the right triangle TPQ we get |TP|^2 + |PQ|^2 = |QT|^2. So
Cz = |TP| = sqrt(|QT|^2 - |PQ|^2) = sqrt( Cx^2 tan(beta)^2 - Cy^2 )
I'm not sure if this is correct but I might as well take a shot. Hopefully I won't get a billion down votes...
I'm too lazy to scale the vectors by the necessary amounts, so I'll assume they are all normalized to have a length of 1. You can make some simple modifications to the calculation to account for the varying sizes. Also, I'll use * to represent the dot product.
A = (1, 0, 0)
B = (cos(g), sin(g), 0)
C = (Cx, Cy, Cz)
A * C = cos(beta) //This is just a definition of the dot product. I'm assuming that the magnitudes are 1, so I can skip that portion, and you said that beta was the angle between A and C.
A * C = Cx //I did this by multiplying each corresponding value, and the Cy and Cz were ignored because they were being multiplied by 0
cos(beta) = Cx //Combine the previous two equations
B * C = cos(alpha)
B * C = Cx*cos(g) + Cy*sin(g) = cos(beta) * cos(g) + Cy*sin(g)
(cos(alpha) - cos(beta) * cos(g))/(sin(g)) = Cy
To be honest, I'm not sure how to get the z component of vector C, but I would expect it to be one more relatively easy step. If I can figure it out, I'll edit my post.
I know the coordinates of A, B and C.. I also know of a vector V originating from C..
I know that the vector intersects A and B, I just don't know how to find i.
Can anyone explain the steps involved in solving this problem?
Thanks alot.
http://img34.imageshack.us/img34/941/triangleprob.png
If you know A and B, you know equation for the line AB, and you said you know V, so you can form the equation for Line V.... Well i is only point that satisfies both those equations.
Equation for Line AB:
(bx-ax)(Y-ay) = (by-ay)(X-ax)
If you knpow the direction (or slope = m) of the vector, and any point that lies on the vector, then the equation of the line for vector V is
Y = mX = b
where m is the slope or direction of the line, and b is the y coordinate where it crosses thevertical y=axis (where X = 0)
if you know a point on the line (i.e., C = (s, t) then you solve for b by:
t = ms + b ==> b = t - ms,
so equation becomes
Y = mX + t-ms
i = C+kV
Lets call N the normal to the line A,B so N = [-(B-A).y, (B-A).x]
Also, for any point on the line:
(P-A)*N = 0 -- substitute from line 1 above:
(C+kV-A)*N = 0
(kV+C-A)*N = 0
kV*N + (C-A)*N = 0
kV*N = (A-C)*N
k = [(A-C)*N]/V*N
Now that we have k, plug it into line 1 above to get i.
Here I'm using * to represent dot product so expanding to regular multiplication:
k = ((A.x-C.x)*-(B.y-A.y) + (A.y-C.y)*(B.x-A.x)) / (V.x*-(B.y-A.y) + V.x*(B.x-A.x))
I.x = C.x + k*V.x
I.y = C.y + k*V.y
Unless I screwed something up....
Simple algebra. The hard part is often just writing down the basic equations, but once written down, the rest is easy.
Can you define a line that emanates from the point C = [c_x,c_y], and points along the vector V = [v_x,v_y]? A nice way to represent such a line is to use a parametric representation. Thus,
V(t) = C + t*V
In terms of the vector elements, we have it as
V(t) = [c_x + t*v_x, c_y + t*v_y]
Look at how this works. When t = 0, we get the point C back, but for any other value of t, we get some other point on the line.
How about the line segment that passes through A and B? One way to solve this problem would be to define a second line parametrically in the same fashion. Then solve for a system of two equations in two unknowns to find the intersection.
An easier approach is to look at the normal vector to the line segment AB. That vector is given as
N = [b_y - a_y , a_x - b_x]/sqrt((b_x - a_x)^2 + (b_y - a_y)^2)
Note that N is defined here to have a unit norm.
So now, when do we know if a point happens to lie along the line that connects A and B? This is easy now. That will happen when the dot product defined below is exactly zero.
dot(N,V(t) - A) = 0
Expand this, and solve for the parameter t. We can write it down using dot products.
t = dot(N,A-C)/dot(N,V)
Or, if you prefer,
t = (N_x*(a_x - c_x) + N_y*(a_y - c_y)) / (N_x*v_x + N_y*v_y))
And once we have t, substitute into the expression above for V(t). Lets see all of this work in practice. I'll pick some points A,B,C and a vector V.
A = [7, 3]
B = [2, 5]
C = [1, 0]
V = [1, 1]
Our normal vector N, after normalization, will look something like
N = [0.371390676354104, 0.928476690885259]
The line parameter, t, is then
t = 3.85714285714286
And we find the point of intersection as
C + t*V = [4.85714285714286, 3.85714285714286]
If you plot the points on a piece of paper it should all fit together, and all in only a few simple expressions.
At present I have a control to which I need to add the facility to apply various acuteness (or sensitivity). The problem is best illustrated as an image:
Graph http://img87.imageshack.us/img87/7886/control.png
As you can see, I have X and Y axess that both have arbitrary limits of 100 - that should suffice for this explanation. At present, my control is the red line (linear behaviour), but I would like to add the ability for the other 3 curves (or more) i.e. if a control is more sensitive then a setting will ignore the linear setting and go for one of the three lines. The starting point will always be 0, and the end point will always be 100.
I know that an exponential is too steep, but can't seem to figure a way forward. Any suggestions please?
The curves you have illustrated look a lot like gamma correction curves. The idea there is that the minimum and maximum of the range stays the same as the input, but the middle is bent like you have in your graphs (which I might note is not the circular arc which you would get from the cosine implementation).
Graphically, it looks like this:
(source: wikimedia.org)
So, with that as the inspiration, here's the math...
If your x values ranged from 0 to 1, the function is rather simple:
y = f(x, gamma) = x ^ gamma
Add an xmax value for scaling (i.e. x = 0 to 100), and the function becomes:
y = f(x, gamma) = ((x / xmax) ^ gamma) * xmax
or alternatively:
y = f(x, gamma) = (x ^ gamma) / (xmax ^ (gamma - 1))
You can take this a step further if you want to add a non-zero xmin.
When gamma is 1, the line is always perfectly linear (y = x). If x is less than 1, your curve bends upward. If x is greater than 1, your curve bends downward. The reciprocal value of gamma will convert the value back to the original (x = f(y, 1/g) = f(f(x, g), 1/g).
Just adjust the value of gamma according to your own taste and application needs. Since you're wanting to give the user multiple options for "sensitivity enhancement", you may want to give your users choices on a linear scale, say ranging from -4 (least sensitive) to 0 (no change) to 4 (most sensitive), and scale your internal gamma values with a power function. In other words, give the user choices of (-4, -3, -2, -1, 0, 1, 2, 3, 4), but translate that to gamma values of (5.06, 3.38, 2.25, 1.50, 1.00, 0.67, 0.44, 0.30, 0.20).
Coding that in C# might look something like this:
public class SensitivityAdjuster {
public SensitivityAdjuster() { }
public SensitivityAdjuster(int level) {
SetSensitivityLevel(level);
}
private double _Gamma = 1.0;
public void SetSensitivityLevel(int level) {
_Gamma = Math.Pow(1.5, level);
}
public double Adjust(double x) {
return (Math.Pow((x / 100), _Gamma) * 100);
}
}
To use it, create a new SensitivityAdjuster, set the sensitivity level according to user preferences (either using the constructor or the method, and -4 to 4 would probably be reasonable level values) and call Adjust(x) to get the adjusted output value. If you wanted a wider or narrower range of reasonable levels, you would reduce or increase that 1.5 value in the SetSensitivityLevels method. And of course the 100 represents your maximum x value.
I propose a simple formula, that (I believe) captures your requirement. In order to have a full "quarter circle", which is your extreme case, you would use (1-cos((x*pi)/(2*100)))*100.
What I suggest is that you take a weighted average between y=x and y=(1-cos((x*pi)/(2*100)))*100. For example, to have very close to linear (99% linear), take:
y = 0.99*x + 0.01*[(1-cos((x*pi)/(2*100)))*100]
Or more generally, say the level of linearity is L, and it's in the interval [0, 1], your formula will be:
y = L*x + (1-L)*[(1-cos((x*pi)/(2*100)))*100]
EDIT: I changed cos(x/100) to cos((x*pi)/(2*100)), because for the cos result to be in the range [1,0] X should be in the range of [0,pi/2] and not [0,1], sorry for the initial mistake.
You're probably looking for something like polynomial interpolation. A quadratic/cubic/quartic interpolation ought to give you the sorts of curves you show in the question. The differences between the three curves you show could probably be achieved just by adjusting the coefficients (which indirectly determine steepness).
The graph of y = x^p for x from 0 to 1 will do what you want as you vary p from 1 (which will give the red line) upwards. As p increases the curve will be 'pushed in' more and more. p doesn't have to be an integer.
(You'll have to scale to get 0 to 100 but I'm sure you can work that out)
I vote for Rax Olgud's general idea, with one modification:
y = alpha * x + (1-alpha)*(f(x/100)*100)
alt text http://www4c.wolframalpha.com/Calculate/MSP/MSP4501967d41e1aga1b3i00004bdeci2b6be2a59b?MSPStoreType=image/gif&s=6
where f(0) = 0, f(1) = 1, f(x) is superlinear, but I don't know where this "quarter circle" idea came from or why 1-cos(x) would be a good choice.
I'd suggest f(x) = xk where k = 2, 3, 4, 5, whatever gives you the desired degre of steepness for &alpha = 0. Pick a value for k as a fixed number, then vary α to choose your particular curve.
For problems like this, I will often get a few points from a curve and throw it through a curve fitting program. There are a bunch of them out there. Here's one with a 7-day free trial.
I've learned a lot by trying different models. Often you can get a pretty simple expression to come close to your curve.