/sci/ - Science & Math

[Anonymous No.5541158]

Hey /sci/ for some project we need to build a ski-jump to launch a round bullet as far as possible

when you know the speed when it leaves the ramp it's easy to get to the distance it travels,
but i'm having trouble finding a good formula to calculate the speed the bullet has right before it leaves the jump.
(i suppose an angle of 45° degrees is optimal)

[other factors is that its a inox/rustfree steel bullet with 1.5 cm diametres (0.6 inch) and that the ramp is 100cm (40inch) hifh max]

[Anonymous No.5541162]

high*

bump

[Anonymous No.5541314]

guise?

bump

[Anonymous No.5541370]

Speed depends on which direction you measure it. Also, 45 degrees is not optimal.

[Anonymous No.5541387]

>>5541370
why isn't 45° optimal ?
at least for a throw it is optimal, why not for a jump?

speed in the direction it leaves the ramp/jump

[Anonymous No.5541395]

For the falling part when you're gaining speed, as close to vertical as possible would be optimal.

[Anonymous No.5541404]

>>5541370
It is actually.

[Anonymous No.5541423]

>>5541395
The angle doesn't matter when you're ignoring friction, and friction would be negligible.

[Anonymous No.5541431]

>>5541423
>ignoring friction

fucking physicists. You always do this shit

[Anonymous No.5541432]

>>5541404
>what is air resistance.

[Anonymous No.5541442]

>>5541432
insignificant.

[Anonymous No.5541451]

>>5541442
which is why all our transportation systems are so efectiont, and bulets traval infinitekly far and termenal velosity is neer the speed of light, wher wefegt all teh hoddpo.

[Anonymous No.5541456]

>>5541158
I think the easiest way to calculate the momentum it has when leaving the ramp is using the conservation of energy. mΔhg=1/2mv2
m, Δh and g are given, so you can solve for v

[Anonymous No.5541462]

>>5541451
The diameter of the ball only has 2 significant digits and it's traviling relativly slowly, so in this case air resistance would be insignificant

[Anonymous No.5541474]

>>5541432
I just did a back-of-the-envelope derivation figuring in air resistance in the horizontal direction and got an optimal take-off angle of 45 degrees. Maybe I should have applied air resistance in the vertical direction, but I was lazy when I got to the second half, plus I figure the y-direction velocity is damped by gravity so air resistance is less significant.

[Anonymous No.5541514]

We're probably going to ignore the air friction since the ball is so small
but since the ball rolls(/or glides) on the ramp surface, I don't think it will be a good idea to ignore friction

[Anonymous No.5541540]

IMO you're going to do better making sure the direction of your motion leading to release is focused than you will focusing more intensely on reducing friction from the launch device.

I'm not saying don't try and minimize friction, but focus on making sure that you go in one direction down the slide so that you don't lose force at launch to side wasted force input.

people are going to go for eliminating friction and ultimately that's much higher hanging fruit.

[Anonymous No.5541579]

>>5541540
first priority is certainly the shape of the ramp
but 1/3 of the points during the evaluation is which ball of all groups flies the farthest

[Anonymous No.5541611]

this is so cool, i wish my school would have had physics project that are so awesome

[Anonymous No.5541622]

>>5541611
Really? Cause my physics class in High School building catapults and predicting where they'd hit. It was good times.

[Anonymous No.5541641]

>>5541622
No, but from what I read here a lot of you guys seem to have stuff like that. Are you American? I have a feeling the US are more hands-on in their education than most of Europe ... and also put more focus on having students present stuff.

My education was mostly learning lots of stuff, little questioning things or experimentation. Though I really can't compain about my physics teacher, but you know. The system and whatnot. Things just don't work that way around here.

[Anonymous No.5541722]

>>5541456
I managed to miss the only really helping reply, thanks
>>5541641
OP here, I'm from Belgium and this project is actually part of my first bachelor in industrial engineering : construction
but in highschool we had similar projects, although there wasn't really a big focus on presentation

[Anonymous No.5541729]

>>5541722
okay then, dunno why my physics classes were so lame-o.

[Anonymous No.5542041]

bump

[Anonymous No.5542097]

Use energy analysis?

[Anonymous No.5542213]

>>5541158
An easy way to get a rough estimate of the launch speed is to assume that all the potential energy at the top of the ramp gets converted into translational energy at the launch height.
If the bullet starts at height <span class="math">x_2[/spoiler] and is launched at height <span class="math">x_1[/spoiler], has a mass <span class="math">m[/spoiler], and the launch velocity is <span class="math">v[/spoiler], then by energy conservation
<div class="math">\frac{mv^2}{2} = mg\left(x_2-x_1\right), </div> where <span class="math">g[/spoiler] is the local gravitational acceleration.
It follows that
<div class="math">v = \sqrt{2g\left(x_2-x_1\right)}, </div> and this is a first approximation of the launch speed. In practice, energy will of course be dissipated into heat/sound/... by friction, turbulence, non-rigidness, et cetera. But more importantly, some energy will be converted into rotational energy in the motion, and this is probably the biggest energy "loss". (I didn't list it earlier, because I knew nothing about rotational energy before university, and perhaps you're in the same boat.)
The rotational energy is <span class="math">E_r=\frac{1}{2}I\omega^2[/spoiler], where <span class="math">I[/spoiler] is the mass moment of inertia, and <span class="math">\omega[/spoiler] is the angular velocity. For a ball (solid), <span class="math">I=\frac{2}{5}mr^2[/spoiler], where <span class="math">r[/spoiler] is the radius of the ball. Using <span class="math">v=r\omega[/spoiler], one arrives at
<div class="math">mg\left(x_2-x_2\right) = \frac{mv^2}{2} + \frac{I\omega^2}{2} = \frac{7}{10}mv^2,</div>
which gives
<div class="math">v=\sqrt{\frac{10}{7}g\left(x_2-x_1\right)}.</div> <span class="math">\Bigl([/spoiler]Note that <span class="math">\sqrt{\frac{10}{7}}\approx 1.2,[/spoiler] and <span class="math">\sqrt{2}\approx 1.4[/spoiler], so the correction is rather big.<span class="math">\Bigr)[/spoiler]

[Anonymous No.5542230]

>>5542213
thanks bro
this looks useful

[Anonymous No.5542236]

>>5542213
I seem to have broken one of my LaTeX-lines. The formula should be
<span class="math">v=\sqrt{\frac{10}{7}g\left(x_2-x_1\right)}.[/spoiler]
You probably will want to check that my calculations are correct, though.

[Anonymous No.5542251]

>>5542236
Strange. It passes the LaTeX-preview, but still doesn't work. The missing line should be sqrt[10/7*g*(x_2-x_1)].

[Anonymous No.5542789]

>>5542213
You can use v=rw only if the ball rolls on the ramp without sliding, which, I think, won't really happen with a metal ball

[Anonymous No.5543447]

Ok I dont know if this has been mentioned cause tldr, but I suppose you could do it based on potential energy (mgh1 - mgh2 - friction etc). The friction is the part that accounts for the angles and stuff but you can just approximate it.

[Anonymous No.5543450]

>>5542097
well fuck its already been said. well there you go.
>>5542213
>>5543447

[Anonymous No.5545405]

unnecessary bump

[Anonymous No.5545908]

>>5541158
I'm going to suggest that you focus on choosing a good launch angle. Not necessarily because it is going to be the most important factor, but because it's easy to control and to experiment with.
I'll go on a limb and claim that you essentially have two options to test what launch angle to use:
1) You have a lot of time in the laboratory (and are willing to take the time), and have a decent amount of material to work with. Set up some test ramp and vary the launch angle. If you have big control over your ramp design, choose a design that you think is going to work well. (Above all, avoid too sharp curvature in the ramp, especially close to the launch. Problems arise especially easily if you are lazy when varying the launch angle by having an abrupt change of direction just before launch.) The speed during travel is going to affect the drag on the bullet, so different launch angles could be more well-suited to different launch speeds. If you have a somewhat fixed ramp design, try varying the launch angle systematically, and measure the length traveled.
2) You don't have a lot of time in the lab, and have to prepare as much as possible beforehand. Do some numerical calculations to determine the optimal launch angle. This will probably be less than <span class="math">45\:^{\circ}[/spoiler] due to drag. The drag can probably be modelled as quadratic, that is <div class="math">\mathbf{F}_D = -C\mathbf{v}^2 \hat{\mathbf{v}},</div> where <span class="math">\hat{\mathbf{v}}[/spoiler] is the normalized velocity vector, which is really just the instantaneous direction of travel, and <span class="math">C[/spoiler] is a drag coefficient which depends on both the bullet and the medium around it. The drag equation can also be stated as <div class="math">\mathbf{F}_D=-C\left| \mathbf{v} \right| \mathbf{v}.</div> You have to determine <span class="math">C[/spoiler] experimentally. This can be done by dropping the bullet, since drag can be solved analytically in one dimension. I'd like to elaborate, but I really have to sleep. Perhaps later.