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August 7th, 2016, 09:40 AM   #1
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Determining the Components of a Velocity Vector

Hi there...

Short and sweet:

I need to figure out a way to do this, but in 3 dimensions. I know there would be a 2nd angle needed (pitch *and* heading).

In a nutshell, I want to be able to simulate a cannon and get the initial velocity vector as soon as the gunpowder ignites.

There are tens of thousands of examples in 2D, but... as soon as you start searching for 3D solutions it gets pretty complicated pretty fast and there is no clear-cut way that I have found.

It seems pretty trivial, right? Or am I being a tad cynical?

Thank you!
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August 7th, 2016, 02:45 PM   #2
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Your problem is 2 dimensional. The plane of interest is through the cannon perpendicular to the ground. Get the vector components in that plane. The third component is 0.
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August 8th, 2016, 09:00 AM   #3
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Mathman is correct. However, if your x-axis is not pointing along the initial velocity vector, you can consider the solution to have two steps, which I shall detail below.

Description of coordinate system:
Cartesian $\displaystyle x,y$ plane with $\displaystyle z$ vertically upwards. Therefore, the initial velocity vector, $\displaystyle \vec{v}$, is of the form ($\displaystyle v_x,v_y,v_z$) and has a magnitude of $\displaystyle v$.

In other words, this is just a bog-standard flat 3D space where x and y are horizontal directions, "forward" and "left", and z is "up".

Known values:
Let $\displaystyle \alpha$ be the horizontal angle anti-clockwise from the x-axis along the $\displaystyle x,y$ plane (i.e. how much you rotate "left"). This can be any number between 0 and 360 degrees.

Let $\displaystyle \beta$ be the vertical angle upwards from the $\displaystyle x,y$ plane (i.e. how much you look "up"). This can be any number between 0 (not looking up) and 90 degrees (looking directly upwards at the sky)

Let $\displaystyle v$ be the initial speed the projectile is expelled (in m/s).


1. Perform trigonometry in the vertical plane along the initial velocity vector to get the $\displaystyle z-$ component of the initial velocity.

$\displaystyle v_z = v \sin \beta$

We can calculate the component in the $\displaystyle x,y$ plane. Let's call it $\displaystyle d$.
$\displaystyle d = v \cos \beta$

2. perform trigonometry in the horizontal plane to get the $\displaystyle x-$ and $\displaystyle y-$ components of the initial velocity.

$\displaystyle v_y = d \sin \alpha$

$\displaystyle v_x = d \cos \alpha$

Your answer is then $\displaystyle \vec{v} = v_x i + v_y j + v_z k$ or ($\displaystyle v_x, v_y, v_z$), whichever you prefer.

This diagram I found on the net should help visualize it, but just be careful because I used z for vertically upwards and y for left, whereas the diagram has y for vertically upwards instead and z for left.

$\displaystyle d$ is basically the red line.

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