I've got a camera attached to a parent that scales causing the camera to "zoom". I want the camera to tilt more at a lower scale. I need an equation that will tilt the camera between the min and max based on the scale of the parent.
Any help would be greatly appreciated =)
See the diagram below:
Instead of scale, you need distances. Consider the variable verical distance y and the target horizontal distance x which you want to keep fixed. The angle of the camera θ is related by
θ = ATAN(y/x)*(180/π)
Given the end conditions y_1/x = TAN(20°) and y_2/x = TAN(40°) one finds that
y_2 = TAN(40°)/TAN(20°)*y_1 = 2.3054*y_1
x = COS(20°)/SIN(20°)*y_1 = 2.7474*y_1
The initial height y_1 is required to compute the horizontal distance x.
Now since s=0.1 means y(s)=y_1 and s=1.0 means y(s)=y_2 then
y(s) = 10/9*(y_2-y_1)*s+(10*y_1-y_2)/9
= y_1*10*(1-s)/9+y_1*(10*s-1)*TAN(40°)/(9*TAN(20°))
= y_1*(1.450*s+0.855)
TAN(θ) = y(s)/x
TAN(θ) = 10*(1-s)*TAN(20°)/9+(10*s-1)*TAN(40°)/9
Use this:
θ(s) = 180/π*ATAN(0.5279*s+0.3112)
With the following example values
s θ(s)
0.1 20°
0.55 31°
1.0 40°
If I'm reading it right, the Scale varies from 0.1 to 1.0, and you want the Angle to vary from 20 to 40 degrees. Right?
A simple linear formula would look like
CurrentAngle = MinAngle + (CurrentScale - MinScale) * (MaxAngle - MinAngle) / (MaxScale - MinScale)
= 20 + (CurrentScale - 0.1) * (40 - 20) / (1 - 0.1)
= 20 + (CurrentScale - 0.1) * 20 / 0.9
So if you use 0.64 as the CurrentScale, as in your example above, you'd get
= 20 + (0.64 - 0.1) * 20 / 0.9
= 32
Linear is the simplest mathematically, but if your application is animated or needs to change the angle faster on one end or the other of your scale, you may get a more polished result from using a formula with a curve to it (logarithmic, parabolic or exponential, maybe?).
Related
I am working on an algorithm which requires the coordinates for the center point of an offset arc. Dimensions available to find this are shown in the image below and the dimension required is labeled as X:
The point at the very left of the arc is a quadrant therefore, the arc center also 2.27 away from the line with a dimension of 3. The actual value of x in this example is 2.47 (rounded), this was found using CAD however, I am trying to find an mathematical relation for this.
Also, please note that dimensions shown are just what are available and not necessarily what are needed to find the center point.
I consider that top arc end is vertical (adjoint with straight line).
Let's denote segments
2.27 = B
4.5-3=1.5 = A
x = R = A + t //arc radius
R^2 = B^2 + t^2 //Pythagor's rule
(A + t)^2 = B^2 + t^2
t = (B^2 - A^2)/ (2*A)
t = 0.967 here
x = R = A + t = 1.5 + 0.967 = 2.467
end formula:
x = A + (B^2 - A^2)/ (2*A) = (A^2 + B^2)/ (2*A)
I've been doing a lot of research on the topic and found a couple of post that where helpful but I just can't get this right.
I am developing a very simple structural analysis app. In this app I need to display a graph showing the internal stress of the beam. The graph is obtained by the formula:
y = (100 * X / 2) * (L - X)
where L is the known length of the beam (lets say its 1 for simplicity). And X is a value between 0 and the Length of be beam. So the final formula would be:
y = (100 * X / 2) * (1 - x) where 0 < X < 1.
Assuming my start and end points are P0 = (0,0) and P2 = (1,0). How can I obtain P2 (control point)?? I have been searching in the Wikipedia page but I am unsure how to obtain the control point from the quadratic bezier curve formula:
B(t) = (1 - t)^2 * P0 + 2*(1 - t)*t * P1 + t^2 * P2
I'm sure it must be such an easy problem to fix… Can anyone help me out?
P.S.: I also found this, How to find the mathematical function defining a bezier curve, which seems to explain how to do the opposite of what I am trying to achieve. I just can't figure out how to turn it around.
We want the quadratic curve defined by y to match the quadratic Bezier curve
defined by B(t).
Among the many points that must match is the peak which occurs at x =
0.5. When x = 0.5,
y = (100 * x / 2) * (1 - x)
100 1 25
y = ---- * --- = ---- = 12.5
4 2 2
Therefore, let's arrange for B(0.5) = (0.5, 12.5):
B(t) = (1-t)^2*(0,0) + 2*(1-t)*t*(Px, Py) + t^2*(1,0)
(0.5, 12.5) = B(0.5) = (0,0) + 2*(0.5)*(0.5)*(Px, Py) + (0.25)*(1,0)
0.5 = 0.5 * Px + 0.25
12.5 = 0.5 * Py
Solving for Px and Py, we get
(Px, Py) = (0.5, 25)
And here is visual confirmation (in Python) that we've found the right point:
# test.py
import matplotlib.pyplot as plt
import numpy as np
x = np.linspace(0, 1, 100)
y = (100*x/2)*(1-x)
t = np.linspace(0, 1, 100)
P0 = np.array([0,0])
P1 = np.array([0.5,25])
P2 = np.array([1,0])
B = ((1-t)**2)[:,np.newaxis]*P0 + 2*((1-t)*t)[:,np.newaxis]*P1 + (t**2)[:,np.newaxis]*P2
plt.plot(x, y)
plt.plot(B[:,0], B[:,1])
plt.show()
Running python test.py, we see the two curves overlap:
How did I know to choose t = 0.5 as the parameter value when B(t) reaches its maximum height?
Well, it was mainly based on intuition, but here is a more formal way to prove it:
The y-component of B'(t) equals 0 when B(t) reaches its maximum height. So, taking the derivative of B(t), we see
0 = 2*(1-2t)*Py
t = 0.5 or Py = 0
If Py = 0 then B(t) is a horizontal line from (0,0) to (1,0). Rejecting this degenerate case, we see B(t) reaches its maximum height when t = 0.5.
Your quadratic bezier curve formula has a typo in the middle term. It should be:
B(t) = (1 - t)^2 * P0 + 2 * (1 - t) * t * P1 + t^2 * P2
This means you should take the P1=(1,50) that #unutbu found and divide the coordinates in half to get P1=(.5,25). (This won't matter if you're plotting the parametric equation on your own, but if you want something like LaTeX's \qbezier(0,0)(.5,25)(1,0), then you'll need the corrected point.)
The P1 control point is defined so that the tangent lines at P0 and P2 intersect at P1. Which means that if (P1)x=(P2)x, the graph should be vertical on its righthand side (which you don't want).
In response to your comment, if you have a quadratic y=f(x), then it is symmetrical about its axis (almost tautologically). So the maximum/minimum will occur at the average of the roots (as well as the control point).
What is best way or an algorithm for generating a random 3d point [x,y,z] inside the volume of the circular cylinder if radius r and height h of the cylinder are given?
How about -- in Python pseudocode, letting R be the radius and H be the height:
s = random.uniform(0, 1)
theta = random.uniform(0, 2*pi)
z = random.uniform(0, H)
r = sqrt(s)*R
x = r * cos(theta)
y = r * sin(theta)
z = z # .. for symmetry :-)
The problem with simply taking x = r * cos(angle) and y = r * sin(angle) is that then when r is small, i.e. at the centre of the circle, a tiny change in r doesn't change the x and y positions very much. IOW, it leads to a nonuniform distribution in Cartesian coordinates, and the points get concentrated toward the centre of the circle. Taking the square root corrects this, at least if I've done my arithmetic correctly.
[Ah, it looks like the sqrt was right.]
(Note that I assumed without thinking about it that the cylinder is aligned with the z-axis and the cylinder centre is located at (0,0,H/2). It'd be less arbitrary to set (0,0,0) at the cylinder centre, in which case z should be chosen to be between -H/2 and H/2, not 0,H.)
Generate a random point inside the rectangular solid circumscribing the cylinder; if it's inside the cylinder (probability pi/4), keep it, otherwise discard it and try again.
Generate a random angle (optionally less than 2π), a random r less than the radius, and a random z less than the height.
x = r * cos(angle)
y = r * sin(angle)
The z axis is easy: -0.5 * h <= z <= 0.5 * h
The x and y are equal to a circle will be:
x^2 + y^2 <= r^2
Buth math is long ago for me :-)
Let's say we have a 100x100 coordinate system, like the one below. 0,0 is its left-top corner, 50,50 is its center point, 100,100 is its bottom right corner, etc.
Now we need to draw a line from the center outwards. We know the angle of the line, but need to calculate the coordinates of its end point. What do you think would be the best way to do it?
For example, if the angle of the line is 45 degrees, its end point coordinates would be roughly 75,15.
You need to use the trigonometric functions sin and cos.
Something like this:
theta = 45
// theta = pi * theta / 180 // convert to radians.
radius = 50
centerX = 50
centerY = 50
p.x = centerX + radius * cos(theta)
p.y = centerY - radius * sin(theta)
Keep in mind that most implementations assume that you're working with radians and have positive y pointing upwards.
Use the unit circle to calculate X and Y, but because your radius is 50, multiply by 50
http://en.wikipedia.org/wiki/Unit_circle
Add the offset (50,50) and bob's your uncle
X = 50 + (cos(45) * 50) ~ 85,36
Y = 50 - (sin(45) * 50) ~ 14,65
The above happens to be 45 degrees.
EDIT: just saw the Y axis is inverted
First you would want to calculate the X and Y coordinates as if the circle were the unit circle (radius 1). The X coordinate of a given angle is given by cos(angle), and the Y coordinate is given by sin(angle). Most implementations of sin and cos take their inputs in radians, so a conversion is necessary (1 degree = 0.0174532925 radians). Now, since your coordinate system is not in fact the unit circle, you need to multiply the resultant values by the radius of your circle. In this given instance, you would multiply by 50, since your circle extends 50 units in each direction. Finally, using a unit circle coorindate system assumes your circle is centered at the origin (0,0). To account for this, add (or subtract) the offset of your center from your calculated X and Y coordinates. In your scenario, the offset from (0,0) is 50 in the positive X direction, and 50 in the negative Y direction.
For example:
cos(45) = x ~= .707
sin(45) = y ~= .707
.707*50 = 35.35
35.35+50 = 85.35
abs(35.35-50) = 14.65
Thus the coordinates of the ending segment would be (85.35, 14.65).
Note, there is probably a built-in degrees-to-radians function in your language of choice, I provided the unit conversion for reference.
edit: oops, used degrees at first
How can I convert a length into a value in the range -1.0 to 1.0?
Example: my stage is 440px in length and accepts mouse events. I would like to click in the middle of the stage, and rather than an output of X = 220, I'd like it to be X = 0. Similarly, I'd like the real X = 0 to become X = -1.0 and the real X = 440 to become X = 1.0.
I don't have access to the stage, so i can't simply center-register it, which would make this process a lot easier. Also, it's not possible to dynamically change the actual size of my stage, so I'm looking for a formula that will translate the mouse's real X coordinate of the stage to evenly fit within a range from -1 to 1.
-1 + (2/440)*x
where x is the distance
So, to generalize it, if the minimum normalized value is a and the maximum normalized value is b (in your example a = -1.0, b = 1.0 and the maximum possible value is k (in your example k = 440):
a + x*(b-a)/k
where x is >= 0 and <= k
This is essentially two steps:
Center the range on 0, so for example a range from 400 to 800 moves so it's from -200 to 200. Do this by subtracting the center (average) of the min and max of the range
Divide by the absolute value of the range extremes to convert from a -n to n range to a -1 to 1 range. In the -200 to 200 example, you'd divide by 200
Doesn't answer your question, but for future googlers looking for a continuous monotone function that maps all real numbers to (-1, 1), any sigmoid curve will do, such as atan or a logistic curve:
f(x) = atan(x) / (pi/2)
f(x) = 2/(1+e-x) - 1
(x - 220) / 220 = new X
Is that what you're looking for?
You need to shift the origin and normalize the range. So the expression becomes
(XCoordinate - 220) / 220.0
handling arbitrary stage widths (no idea if you've got threads to consider, which might require mutexes or similar depending on your language?)
stageWidth = GetStageWidth(); // which may return 440 in your case
clickedX = MouseInput(); // should be 0 to 440
x = -1.0 + 2.0 * (clickedX / stageWidth); // scale to -1.0 to +1.0
you may also want to limit x to the range [-1,1] here?
if ( x < -1 ) x = -1.0;
if ( x > 1 ) x = 1.0;
or provide some kind of feedback/warning/error if its out of bounds (only if it really matters and simply clipping it to the range [-1,1] isn't good enough).
You have an interval [a,b] that you'd like to map to a new interval [c,d], and a value x in the original coordinates that you'd like to map to y in the new coordinates. Then:
y = c + (x-a)*(c-d)/(b-a)
And for your example with [a,b] = [0,440] and [c,d] = [-1,1], with x=220:
y = -1 + (220-0)*(1 - -1)/(440-0)
= 0
and so forth.
By the way, this works even if x is outside of [a,b]. So as long as you know any two values in both systems, you can convert any value in either direction.