Recent content by Darkbound

Okay, still, what do I do with it? To simplify it a bit: v^2 = (du/dt)^2 * u(t)^2 * c^2 *(3) v = du/dt * u(t) * c * sqrt(3) du/dt*u(t) - v/(c*sqrt(3)) = 0 P.S. How do I activate LaTeX?

vx = du/dt * cos(c * u(t)) - u(t) * c * du/dt * sin(c * u(t)) vy = du/dt * sin(c * u(t)) + u(t) * c * du/dt * cos(c * u(t)) vz = du/dt Then when I square them I am using the (a+-b)^2 formula to unwrap them for vx and vy vx^2 = (du/dt)^2 * cos^2(c * u(t)) - 2*du/dt*cos(c *...

I tried to do that and I ended up with an equation of the sort of du/dt * C = 0, it was a veeeeery long equation in the beginning but all sines/cosines etc got canceled in one way or another. Where C is v/(sqrt(c+2)) if I recall correctly (I don't have the papers infront of me).

So, the equations should become like this: x = u(t) * cos(c*u(t)) y = u(t) * sin(c*u(t)) z = u(t) Is this correct? And now I need another equation for u, which binds u that goes from 0 to 1 with respect to time, which can be from 0 to 10 (for example)? And time I will be calculating based on...

I still don't understand what my equations should look like

I don't want to mix the equations, they should not be dependent on each other (e.g. x on z or y on z or z on x), because I want to use the same approach for different sets of equations. If I take their derivatives as they currently are I will get the same result as yours dx/dt = cos(c*t) -...

The equations of a conical helix (what I am trying to use as my trajectory) are x = t*cos(c*t) y = t*sin(c*t) z = t (taken from here http://www.mathematische-basteleien.de/spiral.htm) Where c is a constant In my calculations I was simply creating t as a vector of values from 0 to 1 and using it...

Yes, that's what I was thinking about now, t is not time its a parameter, I was using it to calculate each of the points of the motion, so its not actual time.

I have attached an image to this post, I hope that this clarifies what I mean. I need to find vx and vy such that they always make the same v vector, as you can see, the vector always has the same length, but vx and vy can change in different moments of time

Im sorry, either I don't understand you, or you don't understand me. Lets say we have a vector V that is the our desired speed of the point when it moves through the given trajectory (regardless of what the trajectory is). This vector has 2 components, Vx and Vy. There are infinite amount of...

No it will not give a constant velocity, that is precisely what i am saying, the velocities along each respecitve axis are NOT constant, but when you use each of these individual velocities to calculate the total speed, the total speed should be constant at all times, so at a given moment in...

Is that even possible? If vx is let's say 0.1 m/s and vy 0.4 m/s at all times, wouldn't that make the trajectory simply a line?

Ok I will try to expain it again. The motion is two (actually three dimensional, but for simplicity I am keeping it to 2 for now). I have x and y equations, these are the first equations that I wrote in my post, the second x, y equations I've taken from the 1D constant acceleration equations...

Velocity is the first derivative of the position... So I should take the first derivatives of the equations of the trajectory that I have and then do 0.5^2 = derivative(x)^2 + derivative(y)^2 and that will be an equation with t as unknown in it solve for t and then plug t's back into the...

I have a problem where I am given a trajectory by x(t), y(t) and I am given a constant speed through the whole trajectory. I need to find vx and vy. Equations that I am given: x(t) = 1.5 + 0.5 * t * cos(8*pi*t) y(t) = 1.5 + 0.5 * t * sin(8*pi*t) v0 = 0.5 What I have tried to do is use the 1D...