Is This Theorem for Calculating 2D Distance Original?

[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 traveled 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 traveled along the X axis, take v times t and then subtract from it the distance traveled 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)

 

[jedishrfu]

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?

 

[Mentallic]

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?

The OP is ignoring gravity.

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

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 traveling 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)