Is there a new equation for projectile motion?

[goochmawn314]

Idk if anyone has figured this out, but I believe I've found a new equation for the length of an arc traveled by, say, a football.

I uploaded a picture of it. I don't have any experimental proof for it, but I used tricky mathematical manipulations with standard projectile equations, the perimeter of an ellipse, and the Pythagorean theorem. Do you think this could get published? Sorry if I sound like a noob

 

[goochmawn314]

Note: pi is pi, g is -9.81 m/s2, vi is the initial velocity, and theta is the launch angle.

Crap. That sin at the very end of the square root should be to the 4th power. Sorry

 

[goochmawn314]

Here is the fixed image.

 

[mfb]

What is the first symbol in the root?

No, there is no new formula to find for arc lengths of parabolas.
A football does not follow an ellipse.

 

[tommyxu3]

The arc length of a parabola is hard to presented as a elementary function.

 

[ZapperZ]

The arc length of a parabola is hard to presented as a elementary function.

Hard or not, Mother Nature doesn't care. Your equation is not describing what you advertised.

Zz.

 

[tommyxu3]

Hard or not, Mother Nature doesn't care.

Yes, we all know there just exists a value meeting it.

Idk if anyone has figured this out, but I believe I've found a new equation for the length of an arc traveled by, say, a football.

I uploaded a picture of it. I don't have any experimental proof for it, but I used tricky mathematical manipulations with standard projectile equations, the perimeter of an ellipse, and the Pythagorean theorem. Do you think this could get published? Sorry if I sound like a noob

Maybe you could tell everyone how you got the results in details to get more respond?

 

[Dale]

A football does not follow an ellipse

It would if you kicked it from the space station. ☺

 

[tommyxu3]

It would if you kicked it from the space station.

Then its velocity should be set, or it may be a hyperbola.

 

[Dale]

There is also the possibility of it crashing to the earth, but I don't think that a human kick could produce enough delta V for either. Maybe Superman.

 

[Dale]

Here is the fixed image.

It doesn't look like this equation has the right form

https://en.m.wikipedia.org/wiki/Parabola#Length_of_an_arc_of_a_parabola

From your description it sounds like you might be mixing equations for parabolas and equations for elipses.

 

[goochmawn314]

It takes quite a while to explain how I arrived at this equation, but I guess I'll explain anyway even though I'm wrong.

The perimeter of an ellipse is given by the equation 2pi times the square root of a^2 + b^2 over 2. Rearranging this and making some substitutions, I can say the perimeter equals pi x c x √2. C in this case is the distance between the furthermost point on the major axis to the furthermost point on the minor axis.

Now, I thought the flight path of a projectile sort of looked like an ellipse, so I substituted equations for a^2 and b^2.

These are x = (vi)^2(sin2x)/g and y = vi^2(sin^2x)/g. I applied the Pythagorean theorem to get c, and plugged it into the original deal with pi x c x root2. I cut it in half, since it's clearly only the top half of the ellipse.

It was tempting to think it was similar to an ellipse since, like a projectile, has a point where its rate of change is 0.

 

[Dale]

So I wouldn't say laughably wrong. It seems like good work and maybe some little mistakes.

The first thing to do is to decide if you want to use a parabola or an ellipse. Most projectile problems use parabolas. This is appropriate if gravity is approximately uniform over the trajectory, but if you want to model intercontinental ballistic missiles or satellites then you may need to use ellipses.

 

[SteamKing]

It takes quite a while to explain how I arrived at this equation, but I guess I'll explain anyway even though I'm wrong.

The perimeter of an ellipse is given by the equation 2pi times the square root of a^2 + b^2 over 2. Rearranging this and making some substitutions, I can say the perimeter equals pi x c x √2. C in this case is the distance between the furthermost point on the major axis to the furthermost point on the minor axis.

This formula for the perimeter of an ellipse is incorrect.

Finding a formula for the perimeter of an ellipse is famous for leading to the discovery of a class of non-elementary functions called elliptic functions. The arc length of an ellipse is calculated by evaluating the incomplete elliptic function of the second kind (there are a lot of different elliptic functions), as shown here:

http://mathworld.wolfram.com/EllipticIntegraloftheSecondKind.html

and

http://mathworld.wolfram.com/Ellipse.html {the bottom of this page shows the formula for the arc length of an ellipse}

This article presents some approximate formulas for the perimeter of an ellipse, which don't require the evaluation of elliptic integrals:

https://en.wikipedia.org/wiki/Ellipse#Circumference

 

[goochmawn314]

I got my formula for the ellipse from the Internet. I know it's not exact, but an approximation rather. I'm only in Calculus I (going into Calc II) so I haven't really studied conic sections (calc-related) yet. Thanks for the info and links though.

 

[mfb]

The approximation is bad once the ellipse deviates significantly from a circle. A football will rarely follow anything close to a (half-)circle.

It would if you kicked it from the space station. ☺

I guess we can neglect the number of footballs kicked from the ISS ;).

 

[SteamKing]

I got my formula for the ellipse from the Internet. I know it's not exact, but an approximation rather. I'm only in Calculus I (going into Calc II) so I haven't really studied conic sections (calc-related) yet. Thanks for the info and links though.

The internet is still a wild place for the unwary. You should check multiple sources.

Anywho, the circumference, C of an ellipse is bounded by the following expression:

4(a² + b²)^1/2 ≤ C ≤ π [2 (a²+b²)]^1/2, a ≥ b