For us engineers - a bit of fun

rcsnydley
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Wed Nov 18, 2009 10:12 pm

I think the real problem is calculating how many beers it would take before someone would actually do this.

x=beers

x(distance of slide)2 + 2x(slope of ramp) + x2(distance to pool) + a six pack = likelihood I would actually do it.

Ric


haoli25
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Wed Nov 18, 2009 10:17 pm

wrench wrote:
..... no calculators were injured in the making of this film.




And what fun is there in that? :laugh:


wrench
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Wed Nov 18, 2009 10:24 pm

Ric's equation is the closest. add tequila and testosterone to the left side, then add a Darwin award to right side, and you got it.


AndyT
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Thu Nov 19, 2009 1:10 am

Distance travelled

The total horizontal distance (d) traveled.

d = \frac{v \cos \theta}{g} \left( v \sin \theta + \sqrt{(v \sin \theta)^2 + 2gy_0} \right)

When the surface the object is launched from and is flying over is flat, the distance traveled is:

d = \frac{v^2 \sin(2 \theta)}{g}

As a special case, the distance is given by

d = \frac{v^2}{g}

when the angle (θ) is 45° and the initial height (y0) is 0.

For explicit derivations of these results, see Range of a projectile.
[edit] Time of flight

The time of flight (t) is the time it takes for the projectile to finish its trajectory.

t = \frac{d}{v \cos\theta} = \frac{v \sin \theta + \sqrt{(v \sin \theta)^2 + 2gy_0}}{g}

As above, this expression can be reduced to

t = \frac{\sqrt{2} \cdot v}{g}

if θ is 45° and y0 is 0.

The above results are found in Range of a projectile.
[edit] Angle of reach

The "angle of reach" (not quite a scientific term) is the angle (θ) at which a projectile must be launched in order to go a distance d, given the initial velocity v.

\sin(2\theta) = \frac{gd}{v^2}

\theta = \frac{1}{2} \arcsin \left( \frac{gd}{v^2} \right)

[edit] Conditions at an arbitrary distance x
[edit] Height at x

The height y of the projectile at distance x is given by

y = y_0 + x \tan \theta - \frac {gx^2}{2(v\cos\theta)^2} .

The third term is the deviation from traveling in a straight line.
[edit] Velocity at x

The magnitude, | v | , of the velocity of the projectile at distance x is given by

| v | = \sqrt{v^2 - 2gx \tan \theta + \left(\frac{gx}{v\cos \theta}\right)^2} .

[edit] Derivation

The magnitude |v| of the velocity is given by

| v | = \sqrt{V_x^2 + V_y^2} ,

where Vx and Vy are the instantaneous velocities in the x- and y-directions, respectively.

Here the x-velocity remains constant; it is always equal to v cos θ.

The y-velocity can be found using the formula

vf = vi + at

by setting vi = v sin θ, a = g, and t = \frac{x}{v \cos \theta}. (The latter is found by taking x = (v cos θ) t and solving for t.) Then,

V_y = v \sin \theta - \frac{gx}{v \cos \theta}

and

| v | = \sqrt{(v \cos \theta)^2 + \left(v \sin \theta - \frac{gx}{v \cos \theta} \right)^2} .

The formula above is found by simplifying.
[edit] Angle θ required to hit coordinate (x,y)
Vacuum trajectory of a projectile for different launch angles

To hit a target at range x and altitude y when fired from (0,0) and with initial velocity v the required angle(s) of launch θ are:

\theta = \tan^{-1}{\left(\frac{v^2\pm\sqrt{v^4-g(gx^2+2yv^2)}}{gx}\right)}

Each root of the equation corresponds to the two possible launch angles so long as both roots aren't imaginary, in which case the initial velocity is not great enough to reach the point (x,y) you have selected. The greatest feature of this formula is that it allows you to find the angle of launch needed without the restriction of y = 0.

Derivation

First, two elementary formulae are called upon relating to projectile motion:

x = v t \cos \theta , t = \frac{x}{v \cos \theta} (1)

y = vt \sin \theta - \frac{1}{2} g t^2 (2)

Solving (1) for t and substituting this expression in (2) gives:

y = x \tan \theta - \frac{gx^2}{2v^2 \cos^2 \theta} (2a)

y = x \tan \theta - \frac{gx^2 \sec^2 \theta}{2v^2} (2b) (Trigonometric identity)

y =x \tan \theta - \frac{gx^2}{2v^2}(1+ \tan^2 \theta) (2c) (Trigonometric identity)

0 = \frac{-gx^2}{2v^2} \tan^2 \theta + x \tan \theta - \frac{gx^2}{2v^2} - y (2d) (Algebra)

Let p = tanθ

0 = \frac{-gx^2}{2v^2} p^2 + xp - \frac{gx^2}{2v^2} - y (2e) (Substitution)

p = {\frac{-x\pm\sqrt{x^2-4(\frac{-gx^2}{2v^2})(\frac{-gx^2}{2v^2}-y)}}{2(\frac{-gx^2}{2v^2}) }} (2f) (Quadratic formula)

p = \frac{v^2\pm\sqrt{v^4-g(gx^2+2yv^2)}}{gx} (2f) (Algebra)

\tan \theta = \frac{v^2\pm\sqrt{v^4-g(gx^2+2yv^2)}}{gx} (2g) (Substitution)

\theta = \tan^{-1}{\left(\frac{v^2\pm\sqrt{v^4-g(gx^2+2yv^2)}}{gx}\right)} (2h) (Algebra)

Also, if instead of a coordinate (x,y) you're interested in hitting a target at distance r and angle of elevation φ (polar coordinates), use the relationships x = rcosφ and y = rsinφ and substitute to get:

\theta = \tan^{-1}{\left(\frac{v^2\pm\sqrt{v^4-g(gr^2\cos^2\phi+2v^2r\sin\phi )}}{gr\cos\phi}\right)}

No, I'm no engineer. I just mine their brains.
http://en.wikipedia.org/wiki/Trajectory_of_a_projectile


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Music Junkie
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Thu Nov 19, 2009 11:35 am

LOBO Andy!!!!!!!!!!!!

That was great!

Glad everyone chimed in. With all of the crazy videos out there, it is obvious that many are fakes. That being said, they are still fun!

Sad part about it all, is that back in the day with the help of a little tequila (o.k. a lot of tequila) a bunch of us in college found the biggest hill we could find and went Ice Blocking. Now that was fun! Picture yourself atop a 50lb block of dry ice and a towel to keep your important parts from freezing and heading down a slope of about 35 degrees or so. Luckily no one was seriously injured (again thanks to the tequila). We tried that a few times along with broom ball (playing hockey with a huge red ball on an ice rink while wearing tennis shoes - yeah buddy!).

I wonder why my body protests so much these days......

:silly:


galsteien
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Thu Nov 19, 2009 12:46 pm

Wot concerns me is the REALLY surprised look on the engineers that he actually made it.

Wot! u mean it worked? cor blimey


haoli25
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Thu Nov 19, 2009 1:23 pm

In the real world, engineers are only surprised when something actually works about 95% of the time. :)


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