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Estimation of reach and velocity (formulas) (Read 916 times)

Teg

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Estimation of reach and velocity (formulas)
Jan 29^th, 2007 at 3:28pm

Well, last Sunday I had too much time, so I did some maths and developed some formulas to estimate your start velocity and your reach.
I did this, because I always wondered how far I get and how fast it is.
When you want to know all this, you only have to know how long your stone is in the air. So one person shoots and a second person stops the time.
Then set in your time in the following formulas.
(t) means stopped time.
All calculations are made without air/wind. The ground is plain.
I made this calculations for 45 degree, straight up in the air, and for all angles.

Straight up in the air:
velocity at the start (in meters per second) = 9.81*(t)/2
maximal height (in meters) = 9.81/4*(t^2)-9.81/8*(t^2)

45 degree:
velocity at the start (in meters per second) = √(2) * 9.81*(t)/2
travelled distance (in meters) = 9.81/2*(t^2)
maximal height (in meters) = 9.81/4*(t^2)-9.81/8*(t^2)

For all angles:
velocity at the start (in meters per second) = 9.81*(t)/(2*sin(α))
travelled distance (in meters) = 9.81*(t^2)/(2*tan(α))
maximal height (in meters) = 9.81/4*(t^2)-9.81/8*(t^2)

If you want to convert "meters per second" in "kilometers per hour" just multiply by 3.6 and you got it.

I hope there aren't any mistakes Wink

Enough physics for today.
Your Teg

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Cliff Descens Offline I love Slinging.org! Posts: 142 Gender:	Re: Estimation of reach and velocity (formulas) Reply #1 - Jan 29^th, 2007 at 5:14pm Hi, I looked at the formulas, and I didn't notice any mistakes, but I simplified one: straight up / 45^o / any angle- max hight: 9.81/2 * (t/2)² or 9.81 * t²/8
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CanDo

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Re: Estimation of reach and velocity (formulas)
Reply #2 - Jan 29^th, 2007 at 7:25pm

I wonder how much of an effect air resistance would have. Any of the more enlightened thinkers have a clue? If using uranium ammo, I imagine it would be negligible Wink

An interesting thing regarding physics which I thought of the other day is this:
Every bit of velocity matters when it comes to energy. Kinetic Energy (the stuff that crushes bones and lungs) = 1/2mv^2. Since velocity is squared, the relationship between KE and V is not geometric. For example:
A glande weighing 10 shot at 10 will have 500 energy.
Add just 1 to the velocity and you end up with 605 energy. Add 1 more to the velocity, and you end up with 720 energy. Of course this can all be looked at conversely - the higher the velocities you are slinging at, the harder it will be to add x amount of v.
You can draw your own conclusions as to how this impacts your outlook on slinging...

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wanderer

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Re: Estimation of reach and velocity (formulas)
Reply #3 - Jan 30^th, 2007 at 11:12am

When I started slinging, I thought neglecting air resistance was probably OK - of course it also makes the math less grim Cheesy

.

Then I started wondering why some of my other guestimates of speed didn't match with this one from flight time.... I did some simulations for a 50g spherical bullet made from various material densities just to get a handle on things, and I found the results quite surprising. These were all calculated with a launch at 45 degrees at 40m/s - I hope they are right Undecided

.

1. No air resistance: 163m
2. ~Ice (1g/cc): 82m
3. Stone (3g/cc): 107m
4. Lead(11.3g/cc): 133m
5. Tungsten(19.6g/cc): 141m

So it's clear that air resistance is pretty important if one is slinging for distance. Even for Tungsten we lose quite a distance. Of course, we can get the 'real' projectiles to go a little further by decreasing the launch angle, but these give the general idea I think.

I think the key factor is the 'terminal velocity' of the bullet ie. the speed it will reach falling just by gravity. If we are slinging at speeds more than about 1/4 to 1/2 of that then we can't neglect the effects of air resistance. The terminal velocities for the slugs above range from 32m/s for 'ice' to 87m/s for tungsten.

Of course, then there's the Magnus effect Roll Eyes

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