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This question might sound a little bit broad, but how can I estimate the clarity of a rgb color?
I'm trying to create a function f(rgb) that returns 0 if the rgb is black (0,0,0), 1 if it is white (1,1,1) and something in between for other rgb colors.
What I'm trying is this:
f(rgb) = min(rgb) + c1r + c2g + c3b
First I take the minimum value of the rgb channels, then I add it with the products between each channel and a predefined constant.
Example: rgb = (0.4, 0.8, 0.5)
f(rgb) = min(0.4, 0.8, 0.5) + 0.4 c1 + 0.8 c2 + 0.5 c3
f(rgb) = 0.4 + 0.4 c1 + 0.8 c2 + 0.5 c3
But obviously I don't know the values of the constants, I would have to test them manually.
This is why I'm asking if there is a way already.
Got an idea: convert the color to grayscale.
Here I could find an algorithm that does that:
f(rgb) = 0.299r + 0.587g + 0.114b
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I am really new to D3 and i m trying to understand how the the <g> are positioned in the svg.
Take the following example:
https://stackblitz.com/edit/angular-packed-bubble-chart?file=src%2Fapp%2Fd3-packed-bubble-chart.service.ts
How would be possible to approach the 2 clusters?
The g elements are positioned in lines 73-75 by setting the g elements' transform attribute to "translate(" + d.x + "," + d.y + ")" via the the accessor function.
Essentially, applying a translate(x,y) value to a transform attribute on an element, means shifting the element x pixels horizontally and y pixels vertically.
Check out the documentation for the transform attribute and what it does with a translate value.
The values of d.x and d.y are computed using D3's pack and hierarchy functions.
Hope this helps!
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Position P is start with 7.56m and velocity V is 6(m/s). Acceleration A is -2 (m/s): How to get P when V becomes 0 and how to recalculate acceleration A for final P can be whole number by rounding off?
use equations of motion.
v2 = u2 + 2as
here,
v=0, u=6m/s and a=-2m/s2
So,
s = 9m
So, final position will be 7.56 + 9 = 16.56m
and since acceleration does not change, final acceleration = initial acceleration = -2m/s2
EDIT:
In that case, again use the same equation. Changed values will be:
v=0, u=6, s=9.44
So,
a = -1.907m/s2
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the question is simple one object is moving from east-west with a velocity of v1 and another from south-north with velocity v2.
I just need the algorithm(formula) to calculate the shortest distance between them so I can write a program for it.
I do have distance between them and the meting point of their paths they are d1 and d2.
Assuming you are asking for 2-d space, at t=0, let the starting points be (d1,0) and (0,d2) on the coordinate axes. We can assume this because one object is always moving horizontally (E-W direction, along X-axis) and other vertically (S-N direction, along Y-axis). Now, after some time t, their positions will be,(d1-t*v1) and (0,d2-t*v2). (Speed-Distance-Time relation).
Now, distance between them at this time t will be,
D = d^2 = (d1-t*v1)^2 + (d2-t*v2)^2
So, differentiating both sides wrt t,
dD/dt = 2(-v1)(d1-t*v1) + 2(-v2)(d2-t*v2) ....(1)
For D to be minimum, dD/dt = 0 and second differential must be positive. Now, second differential :
d2D/dt2 = 2*v1^2 + 2*v2^2 which is positive for all real v1/v2. So, if `dD/dt = 0`, distance will be minimum.
So, equating (1) = 0, we get
t = (d1v1 + d2v2)/(v1^2 + v2^2)
So, get sqrt(D) at t = --above value-- and that shall be your answer.
PS: ask these type of questions on mathematics stackexchange.
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I have a class which creates a sprite at a random point on the screen, this sprite then moves to the center of the screen.
How can I make it so that it always travels at the same speed?
obviously
CCMoveTo* move = [CCMoveTo actionWithDuration:5 position: ccp(screenWidth/2, screenHeight/2)];
Will always mean the duration is 5 seconds regardless of the distance. But I want the speed to be constant if its travelling 50 pixels or 500 pixels.
Any Help much appreciated
Calculate the duration from the distance to the center.
duration = distance / rate;
Say that moving 50 pixels in 5 seconds is okay. Then your rate is 10 pixels/second.
rate = 10;
If your sprite is at (x,y) then distance is by the pythagorean theorem
dx = x - screenWidth / 2;
dy = y - screenHeight / 2;
distance = sqrt(dx * dx + dy * dy);
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I have a question [ silly - I admit ] about finding wherether the point is in the circle or not, I have a coordinates of the center of circle, and I know the equation, but I'm having the problem with radius, let say it is 2 km, so I have x:46.123654 y: 15.789456 and r=2 or 200 or 2000? What should be the value of R?
For clarity, the units of r should almost certainly be the same as the units of x and y. If, for instance x = 46.123654 means 46.123654 meters, and the radius of your circle is two kilometers, then the value of r should be 2000.0, meaning two thousand meters. You should also be explicit in some comment about what the units are, i.e. x = 46.123654 //meters. If the units are the same, you can apply formulas without confusing conversions, for instance:
//determines whether a point (x, y) is in the circle of radius r centered at (0, 0)
bool isInCircle(double x, double y, double r)
{
return x * x + y * y <= r * r; //pythagorian theorem!
}
This isn't really programming, just middle-school math and common sense.