# A (hopefully) original theorem

1. Apr 30, 2015

### SpanishOmelette

Hello fellow members of PF,

I believe I may have created a theorem for calculating the distance travlled on the 2-dimensional plane when gravity and other forces are ignored. Now, if this has been thought of before, that's not my fault. I have not studied this before.

The theorem is this... the distance travelled along the Y axis is equal to the gradient of travel divided by 90, then multiply the speed(m/s) by the time taken (s) (yes, this is d=vt), and then multiply both those terms together.

Then, to also find the distance travelled along the X axis, take v times t and then subtract from it the distance travelled along Y. Sounds fiddley in sentences. But I have enclosed a whiteboard sketch.

So, is this original? Does it work? Has it been thought of before?

Mahmoud.

g/90 x vt = distance moved in y, where g = gradient , angle.

motion along x = (vt) - (g/90 x vt)

2. Apr 30, 2015

### Staff: Mentor

Your problem is that of projectile motion and you can determine if your nethod works by comparing it to how its done in introductory physics:

https://en.m.wikipedia.org/wiki/Projectile_motion

Does your solution when plotted on the xy plane produce a parabolic arc?

3. Apr 30, 2015

### Mentallic

The OP is ignoring gravity.

No, it's wrong.

What you've done is to complicate simple trigonometry.

Given a right triangle that sits on the x-axis and the hypotenuse is the distance (d) and $\theta$ is the angle that you're travelling at (where $\theta=0$ is along the x-axis), then the distances x and y are

$$x=d\cos\theta$$
$$y=d\sin\theta$$

You can also replace d by vt if you wish, since d=vt. Also, $\theta$ is in radians. If you use degrees (g, to use your parameters) instead then you'll have to make the transformation

$$\cos\theta = \cos{\frac{\pi g}{180}}$$

So finally you'll have

$$x=vt\cos{\frac{\pi g}{180}}$$
$$y=vt\sin{\frac{\pi g}{180}}$$

Now, g/90 may be incorrect, but it's not too bad of an approximation (for values between 0-90 degrees, but if you used more than that then it's way off), but a better approximation (which is just approximating sin(x) and cos(x) at $x=\pi/4$ radians, or 45 degrees) would be

$$\sin\frac{\pi g}{180}\approx \frac{1}{\sqrt{2}}\left(1+\pi\left(\frac{g-45}{180}\right)-\pi^2\left(\frac{g-45}{180}\right)^2\right)$$