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y=0 and x=0
Both increment by 1 every millisecond, but x=0 whenever it reaches 120.
What is the equation for finding x if y is... 5000?
x = y % 120
% is called modulus
You just use modulus.
(y / 120) + (y % 120) = x
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This is a follow-up question to this question.
Taking the following image as an example:
What I know:
x and y coordinates of points D, E, and P.
Therefore, I also know slope and intercept of D-E line
What I want to know:
x and y coordinates of point Q. (This is the point which crosses the D-E line).
Notation P=[px,py], D=[dx,dy], E=[ex,ey], Q=[qx,qy]
First:
R=P-D=[px-dx, py-dy]=[rx,ry]
K=E-D=[ex-dx, ey-dy]=[kx, ky]
Then
z=dot(R,K)/dot(K,K)=(rx*kx+ry*ky) / (kx*kx+ky*ky)
Finally
Q=D+z*K=[dx+z*kx, dy+z*ky]
The R is vector which start on point D and ends on point K, the K is vector which start on point D and ends on point E. Using this we made scalar projection to calculate result Q. More info about concept here
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Please forgive my lack of knowledge, which i think it's one of those basic formula related to Trigonometry.
Let's look at visual example:
I have 5 lines, with their line equation (let's say they have zero offset ok)
how can i calculate the Theta of each line equation make (in Pi)?
also I have seen this:
Are they generated from Theta of line equations? or it's another theory which help to find the theta?
much appreciate your time and effort
For equation
y = k * x
tg(Theta) = k
and
Theta = Arctg(k) //arctangent function
General line equation
A * x + B * y + C = 0
(It is more general than y=ax+b and includes cases of vertical and horizontal lines)
Theta = atan2(A, B)
(function atan2 or ArcTan2 exists in math libraries of many programming languages)
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I want to rotate a plane represented by the equation z = 6 , by n degrees along y axis and find the new equation of the plane. how can this be done?
Thanks
Base point (0,0,6) after rotation will lie in XZ plane with coordinates
(x0, y0, z0) = (-6*sin(Fi), 0, 6*cos(Fi))
normal vector
n = (A,B,C) = (-sin(Fi), 0, cos(Fi))
so new plane equation is (for explanation see beginning of the article)
A*(x-x0)+B*(y-y0)+C*(z-z0)=0
or
-sin(Fi)*x + cos(Fi)*z - 6 = 0
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Does anyone know how to reverse this equation when only Y, Z are known? I want to know what X is.
(X + Y) % 62 = Z
(X + Y) % 62 = Z
means that X+Y = 62*n + Z for arbitrary integer n, from that follows
X = 62*n+Z-Y
where n can be any integer value, if you need single solution you can pick arbitrary one or if you have any extra limitations on X value you'll need to find X that satisfies them
X = Z - Y + 62*n
where n is any integer.
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how can we calculate the coordinates of a line (Formula used) when the length and angle is given.
I have to make a pop up box which will take as input the length and angle and will draw a line.
just need the formula
Assuming one end of the line is at (x0, y0), the other end will be at:
x1 = x0 + r cos(t * pi / 180)
y1 = y0 + r sin(t * pi / 180)
where r is the length of the line and t is the angle in degrees.