Funny hammerhead conversation

[yeahthatsme]

Originally posted by rdb123
lamby, this is from my post over at PBN a few months ago:



Just change the values in my initial equation to find what you're looking for. Ironically enough, I wrote that reply to argue why the hammerhead is all hype.

http://pbnation.com/showthread.php?s...1&pagenumber=1 (Read it, it's pretty funny)

that was one of the funniest threads i have ever read, way to stick it to'em guys

[rdb123]

Originally posted by lamby
rdb,

Thank you very interesting... Is there a math formulat that would determain maximum distance though? I know that the peak attack angle would have to be figured out, but that is what I have no idea how to figure out

The angle to get the most displacement along the X-axis would have to be a 45 degree angle. I'll do the kinematic equations on paper later tonight and then type them up here on AO. It's Physics 101 really.

Now, my post I quoted and the answer I will give you later do not take into account wind resistence. However, wind affects objects of less mass than those of greater mass, so figuring in wind resistence with Calculus would be moot.

[Miscue]

Originally posted by rdb123


The angle to get the most displacement along the X-axis would have to be a 45 degree angle. I'll do the kinematic equations on paper later tonight and then type them up here on AO. It's Physics 101 really.

Now, my post I quoted and the answer I will give you later do not take into account wind resistence. However, wind affects objects of less mass than those of greater mass, so figuring in wind resistence with Calculus would be moot.

You're joking right? If you do this w/o taking drag into account, your numbers will be severely exaggerated. Also, it does not work out to 45 degrees when you throw in drag... but it's in the area.

[hitech]

Here is a link to a trajectory calculator:

http://home.comcast.net/~dyrgcmn/pball/trajectory.html

[rdb123]

Originally posted by Miscue


You're joking right? If you do this w/o taking drag into account, your numbers will be severely exaggerated. Also, it does not work out to 45 degrees when you throw in drag... but it's in the area.

Yes I know. From what I understood, lamby wanted to know the theoretical distance if fired in a vacuum. That's what I did all this work for, but I'm sure it's probably wrong. I'm just an AP Physics student be hammered by E&M stuff right now.

Now the final answer to how far a paintball could theoretically travel if fired from ground level:

For projectile motion, we must solve simultaneous equations.

Velocity = v
Acceleration = a
Time = t

Velocity{x} = 300fps * cos(45 degrees) = 212.13fps = 64.66m/s

Velocity{y} = 300fps * sin(45 degrees) = 212.13fps = 64.66m/s

Now, concentrating ONLY on the vertical Y-axis motion of the projectile, we must find how long it takes from the time the paintball is fired till when it lands.

v{final} = v{initial} + at

= 0 = 64.66m/s + (-9.8m/s)t
= -64.66m/s = -9.8t

t = 6.60 seconds for the ball to reach it's maximum height if fired at a 45 degree angle from ground level.

The ball's maximum vertical distance is:

Delta(Y) = v{initial}t + .5at^2

= Delta(Y) = (64.66m/s)(6.60 seconds) + .5(-9.8m/s)(6.60 seconds)^2
= Delta(Y) = 426.76 - 213.44
= Delta(Y) = 213.32m

For the time it takes for the paintball to hit the ground from it's highest point, we use:

Delta(Y) = (acceleration)*(time^2)/2

We want the ball to fall 213.32 meters.

= -213.32 meters = (-9.8m/s)*(time^2)]/2
= -426.64 meters = (-9.8m/s)*(time^2)
= 43.53 meters = time^2
= time = 6.60 seconds

It takes a paintball nearly 6.60 seconds for it to fall if dropped at a height of 213.32 meters.


Now, we add the time it takes for the projectile to reach its maximum height, and the time it takes for the projectile to hit the ground.

T = T{1} + T{2}

The total time for a paintball to be fired at 300fps at a 45 degree angle and land is nearly 13.2 seconds.

Since we now know the total time, we plug that into our other equation for the horizontal X-axis motion of the projectile, and we figure:

Displacement/Delta(X) = v{initial}t + .5at^2

= Delta(X) = (64.66 m/s)t + .5(a)t^2

Since acceleration in the horizontal axis is zero (we are neglecting wind resistence), the equation simplifies to:

Delta(X) = (64.66 m/s)t
= Delta(X) = (64.66 m/s)*(13.2 seconds)

The farthest theoretical distance a paintball fired at a 45 degree angle from groundlevel, travelling at 300fps is 853.51 meters.

NOTE: For some reason this seems wrong. I did the equations in a real hurry. I'd appreciate it if Miscue, manike, or some other physics buff could check over my work and see if I did anything wrong. My conclusion should not be set in stone. Hope that helped some.

Like I said, I most likely messed up somewhere in there. Please don't be too harsh. This stuff was the second week of school. Obviously I need to review before my AP Exam.

[lamby]

rbd, thank you for the eqautions.

I think something is greatly wrong though. I dont think that even in a vaccum that a paintball fired at 300fps with an attack angle of 45deg can reach a distance of 853 meters. That sounds WAY to high. I was thinking around a maximum distance of 75 meters. What I wanted included air resistance accross a perfect sphere of .691 inches (I know no paintballs are not perfect, but the drag coeficient would be easier to calculate), but no wind.

I will look at the eqautions and see if I can remember my basic physics from 17 years ago.

[fallout11]

Given a paintball:
fired at 300 fps
from a height of 5 feet (shoulder height)
at an angle of 45 degrees

will reach a maximum height of 141.6 feet (downrange 184.1 feet, time in flight approx 2.15 seconds)

and a final total distance of 291.8 feet (97.3 yards)
striking the ground at a speed of 65.97 feet per second

total time in flight: 5.75 seconds

It's an enormous "lob", as in archery. Think javelin toss. Same physics.

Rdb123's equations are correct, except they neglect air resistance and drag, which makes a big difference on something as lightweight as a paintball.
In a vacuum, but with earth's gravity, a paintball could indeed fly 800 yards.

[hitech]

Guys, run the calculator I linked to. It is specifically for firing a paintball. The huge results of drag caused by air friction will be very apparent (note how "fast" the paintball slows).

The source code is also available, so you can check the calculations if you like.

[fallout11]

Nice calculator, Hitech!

Yep, it gives pretty much the same numbers I came up with.
Best of all, you don't have to run all the math! Gah!
Ah, well....it was a trip down memory lane.

[fallout11]

Point is, anyone's claim of hitting small targets at 100 yards is total BS.
And I mean TOTAL BS.
You can't argue with mathematics.

Somebody either made a simple typo, or are really laying it on thick.

[rdb123]

Originally posted by hitech
Guys, run the calculator I linked to. It is specifically for firing a paintball. The huge results of drag caused by air friction will be very apparent (note how "fast" the paintball slows).

The source code is also available, so you can check the calculations if you like.

Too bad we couldn't run that calculator using different diameter paintballs. =(

[lamby]

that calculator is very cool (thanks hitech)

That is further than I though a paintball could travel. I never would have though a paintball could travel that far even with a hugh arch. I figured more along the lines of 75-80 yards top.

Very cool discussion, and it goes to prove that there is no way to accurately hit a 11" target at 100 yards with a pb gun.

:) I still think it was a typo :)

Thnaks for all the insight on this subject and for the formulas. I will use these in the future to try to understand what I bearly understood in 9th grade