## Flight distance

I've been having a discussion with some friends at a local Airsoft club for some time now and we are unable to agree on the flight distance's of identical objects that have different masses. I'll try to explain...
If a 6mm spherical object, weighing 0.2g, was fired on a horizontal path with a muzzle velocity of 350 feet per second (we've worked that out as 1.14joules?), given normal atmospheric pressure, gravity and let's say a muzzle height of 60 inches.

And then have a second identical object fired with the same statistics but it weighs 0.25g and fired with the same energy 1.14 joules (we worked out the FPS would be 313).

Now half believe the .25g would travel further than the .2g due to the momentum how ever half believe the .2 would travel further due to the higher velocity?? Can anyone help me with working out how far that object would travel before it hits the ground (resulting bounces are of no interest) or explain which one has the greatest travel distance and why? Many thanks in advance!

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 Mentor Welcome to PF, I'll start by answering the easy question. In a vacuum, the less massive ball travels farther, due to its greater horizontal velocity. The only force acting is gravity, which is in the vertical direction. Therefore, the horizontal momenta are not relevant. Falling objects all accelerate at the same rate (independently of their masses). Therefore, the two balls take the same amount of time to hit the ground, and whichever one is travelling faster horizontally will cover a larger horizontal distance in that time interval. Now, as for the inclusion of air resistance, that's not something I can just answer off the top of my head. I'll have to think about it and look some things up.
 Many thanks and it's making sense so far which is good :) Don't know if it's of any help but this discussion has been going on sometime and there is some info/calculations at http://www.arniesairsoft.co.uk/?filn..._dist_time.htm ignore the bumf at the start and scroll down to the spreadsheet (assuming he has all the correct numbers (sorry not smart enough to work it out myself)) he makes a valid point however I don't see him taking gravity into his calculations and the ball bearings seem to travel forever without dropping out the sky no matter how slow they travel?!? That's said maybe its best to ignore his numbers incase there is another oversight...

Mentor

## Flight distance

Well, the drag force on the objects depend on the density of the fluid (in this case air), the effective area that the object presents to the flow (in this case just its cross-sectional area), and a number called the "drag coefficient" that characterizes how aerodynamic that object is. I'm not sure what you mean by "the same statistics", but if you mean that the objects have the same properties (radius, in particular), then I imagine that their drag coefficients would be the same. Since they have the same radius, their effective areas are the same too.

However, the drag force is the product of all of the above, multiplied by the square of the velocity. In other words, the faster an object is moving through the fluid, the more drag it experiences. That makes the motion kind of hard to solve for. If I ignored drag in the vertical direction as being negligible and only worked out drag in the horizontal direction (in which the speed is much larger), I could probably work it out, but I don't know how much you care. Given that these are bullets, I also don't know how much it will really change the answer to your question.

 If it's too time consuming to work it all out would it be possible to apply the effect of gravity onto the calculations made in the link (above post)? Q: If 2 objects are dropped from the same height with the same aerodynamic properties they hit the ground at the same time right? (lets say 1 second to hit the floor) So if those objects were moving 'forward' would I be correct in saying they would still hit the floor after 1 second flight time, no matter what speed they are travling forward? If that's right would it be as simple as calculating the time it takes to objects to hit the floor (help anyone?) then finding that time on the spreadsheet and seeing which weight ball bearing has travled the furthest?
 Recognitions: Homework Help What you'd have to do is plot velocity versus time for both objects, then calculate the distance which is equal to the area under the curves (integrate the velocity versus time equations) from time the objects exit the muzzle till the time they hit the ground. Although the lighter object has a higher initial velocity, it's slowing down at a faster rate (even when at the same velocity), due to having less mass but the same area (same aerodynamic drag). I don't know which will travel further. Is there anyway to actually test this and measure the average distance, such as an air rifle mounted to a tri-pod and shooting 10 of each type of object?
 Unfortunately a practical test wouldn't be accurate as they have a rubber part to create backspin on the ball bearings giving lift whilst in flight. But I've taken this out of the equation to save complication (it's unpredictable ). You suggest using the 'time the objects exit the muzzle till the time they hit the ground'. Is there any formula to work that out or is it a case of stopwatch and repeatedly dropping a ball from correct height to get a time?? Getting late and I'm getting confused!?! I'll sleep on it and be back fresh tomorrow so you can all melt my brain again!
 Recognitions: Homework Help It will take a bit less than .6 seconds for the pellets to fall from 60 inches. Assuming the spread sheet data is correct, after .3 seconds, the lighter pellet is traveling slower than the heavier pellet, and between .4 and .5 seconds, the distance traveled is about the same, and between .5 and .6 seconds, the heavier pellet has traveled slightly farther. However the spreadsheet initial velocities are different than what was shown the first post of this thread.
 Thanks for that much appreciated. Just noticed the lower initial velocity. So that will have the same effect on both weights, higher speed causing more drag increasing the deceleration. Is it possible/easy to work those numbers back to have a starting velocity of 350fps? Or should I assume the higher muzzle velocity will only give the same results?? Lighter object travelling furthest until .3-.4 seconds only to be overtaken by the heavier. Yet again many thanks for all the help given.

Recognitions:
Homework Help
 Quote by blackhawk So that will have the same effect on both weights, higher speed causing more drag increasing the deceleration.
and both objects having the same drag at the save velocity, but a higher amount of deceleration on the lighter object.

 Is it possible/easy to work those numbers back to have a starting velocity of 350fps?
If the starting velocity is the same for both objects, than other than the initial velocity, the lighter object will have lower velocity due to the higher rate of deceleration at the same velocty but smaller mass. If you have your friend with the spread sheet redo the graphs with these numbers, you'll see a peak difference between the velocities when the rates of deceleration converge to the same number (when vlighter2 / vheavier2 = mlighter / mheavier), then both curves will asymptotically approach zero, but this will occur long after the less than .6 seconds it takes for them to fall 60 inches.

 Sorry my fault I wasn't clear. The lighter object would have a starting velocity of 350 feet per second and the heavier 313 FPS. Trying to use the same power output (1.14joules?). I dont know the person who created those particular graphs and in currently trying to get hold of them, if I do I'll ask if they can rework the numbers.

Mentor
 Quote by blackhawk (assuming he has all the correct numbers (sorry not smart enough to work it out myself)) he makes a valid point however I don't see him taking gravity into his calculations
He doesn't. In fact, he says so in the article:

 Quote by Zinger 084 Although we know that a bb fired from a stock gun at an almost horizontal shot will not stay in the air for 3 seconds, it is important to show the trend involved with this length of time.
So he acknowledges that the BB won't stay in the air for as long as 3 seconds. However, I agree that it was an oversight on his part not to consider the upper limit to how far the BBs could travel due to gravity. As I will show below, it affects some of the conclusions that he makes at the end of the article about which type of BB is better for which range of shot.

 Quote by blackhawk and the ball bearings seem to travel forever without dropping out the sky no matter how slow they travel?!?
I get what you're saying here, but be careful! It's not that the projectile moves in a straight line until it slows down to a certain speed and then starts dropping. It is falling continuously throughout the motion. It starts falling as soon as it leaves the barrel. It can't defy gravity.

 Quote by blackhawk If it's too time consuming to work it all out would it be possible to apply the effect of gravity onto the calculations made in the link (above post)? Q: If 2 objects are dropped from the same height with the same aerodynamic properties they hit the ground at the same time right? (lets say 1 second to hit the floor)
I don't know if that's true in the case where there is drag in the vertical direction. Even if their sizes are the same, I have a feeling the mass might affect the motion. But it is certainly true in the case where they are in free-fall (i.e. gravity only) that they will hit the ground at the same time. In the case of free fall, the time to hit the ground is:

$$t = \sqrt{\frac{2h}{g}}$$

where h is the starting height, and g is the acceleration due to gravity. For a height of 60 inches, that's 0.557 seconds.

 Quote by blackhawk So if those objects were moving 'forward' would I be correct in saying they would still hit the floor after 1 second flight time, no matter what speed they are travling forward? If that's right would it be as simple as calculating the time it takes to objects to hit the floor (help anyone?) then finding that time on the spreadsheet and seeing which weight ball bearing has travled the furthest?
Yes, and that's basically the approach that I took to solving this problem. I assumed that there was drag in the horizontal direction, but no drag in the vertical direction. MAYBE this is a good approximation, I don't know. After all, the drag force is proportional to the speed squared, and the vertical speed is much less than the horizontal speed. So maybe we can neglect vertical air resistance and assume that the BBs are in free fall. Doing that, I got the following result for the horizontal velocity of the projectile vs. time:

$$v_x = \left[\frac{\rho A C_d}{2m} t + \frac{1}{v_{0x}}\right]^{-1}$$

where $\rho$ is the density of air, A is the area, m is the mass, Cd is the drag coefficient, and v0x is the initial (starting) horizontal velocity.

I used that result to figure out what the position of the projectile vs. time would be. It is given by the expression:

$$x =\frac{2m}{\rho A C_d}\left[\ln \left(\frac{\rho A C_d}{2m} v_{0x}t + 1 \right) \right]$$
where ln is the natural logarithm. I have included my derivaton of these results below (at the very end of the post). I don't know if you'll get much out of it, but I've included it for completeness, as well as in the hope that somebody around here will catch any mistakes I might have made.

Plugging in numbers gave me the results that are tabulated below. As you can see, my numbers differ from his quite a bit. That could be because I solved for it exactly, whereas he used some sort of approximation method. Or it could be because what he did is just entirely wrong (or it could be that what I did was wrong).

As you can also see, I've attached two plots, using the same colour convention as him, where the green curve is for the lower mass BB, and the blue curve is for the higher mass one. I've included a vertical grey line that represents the cutoff time of 0.6 seconds (the time required to hit the ground). Everything to the right of this line is irrelevant. As you can see, although the green (less massive) BB starts out with a higher velocity than the blue, it decreases faster, and there is a crossover point. This crossover point occurs BEFORE the cutoff time. Therefore, the more massive BB IS moving faster horizontally when the two projectiles hit the ground.

In contrast, the distance vs. time plot shows that the crossover doesn't happen until after the cutoff time. So, in this case, the less massive (green) BB is always farther along than the more massive (blue) BB, just like in the vacuum case. This means that his conclusion at the end of the article that the more massive BB is better for shots greater than a certain distance is dubious, given that those distances are beyond the range of the rifle in the first place! Clearly it was important to take gravity into account. Of course, if you started with a larger height, the vertical grey line would move to the right, meaning that in some situations it might be possible to get the crossover to occur, and the more massive BB will travel farther.

So, my graphs are qualitatively the same as his (they show the same general trends), but our numbers are not the same. I'm very curious to know how he went about solving the problem. Maybe you could e-mail him with a link to this thread and see what he says?

Velocity vs. time for the 0.20 g BB:
Code:
m = 0.20 g

Time [s]	Velocity [m/s]
0.0		85.000000
0.1		72.024862
0.2		62.486397
0.3		55.178893
0.4		49.401597
0.5		44.719421
0.6		40.847941
0.7		37.593382
0.8		34.819166
0.9		32.426259
1.0		30.341102
1.1		28.507912
1.2		26.883621
1.3		25.434447
1.4		24.133518
1.5		22.959194
1.6		21.893851
1.7		20.922991
1.8		20.034579
1.9		19.218539
2.0		18.466375
2.1		17.770869
2.2		17.125852
2.3		16.526018
2.4		15.966781
2.5		15.444153
2.6		14.954655
2.7		14.495233
2.8		14.063197
2.9		13.656169
3.0		13.272040
Velocity vs. time for the 0.25 g BB:

Code:
m = 0.25 g

Time [s]	Velocity [m/s]
0.0		76.026311
0.1		67.345274
0.2		60.443548
0.3		54.824941
0.4		50.162062
0.5		46.230171
0.6		42.869869
0.7		39.964963
0.8		37.428752
0.9		35.195232
1.0		33.213267
1.1		31.442624
1.2		29.851217
1.3		28.413141
1.4		27.107254
1.5		25.916132
1.6		24.825282
1.7		23.822554
1.8		22.897685
1.9		22.041944
2.0		21.247862
2.1		20.509005
2.2		19.819806
2.3		19.175421
2.4		18.571618
2.5		18.004680
2.6		17.471331
2.7		16.968671
2.8		16.494125
2.9		16.045400
3.0		15.620444
Position vs. Time for the 0.20 g BB:
Code:
m = 0.20 g

Time [s]	Position [m]
0.0		0.0000000
0.1		7.8154560
0.2		14.518450
0.3		20.386575
0.4		25.604955
0.5		30.303238
0.6		34.575765
0.7		38.493329
0.8		42.110412
0.9		45.469844
1.0		48.605903
1.1		51.546450
1.2		54.314438
1.3		56.929004
1.4		59.406264
1.5		61.759923
1.6		64.001732
1.7		66.141840
1.8		68.189076
1.9		70.151167
2.0		72.034912
2.1		73.846329
2.2		75.590768
2.3		77.273005
2.4		78.897324
2.5		80.467581
2.6		81.987259
2.7		83.459514
2.8		84.887218
2.9		86.272987
3.0		87.619215
Position vs. time for the 0.25 g BB:

Code:
m = 0.25 g

Time [s]	Position [m]

0.0		0.0000000
0.1		7.1510455
0.2		13.528054
0.3		19.282345
0.4		24.524790
0.5		29.339053
0.6		33.789830
0.7		37.928175
0.8		41.795090
0.9		45.423999
1.0		48.842510
1.1		52.073688
1.2		55.137002
1.3		58.049037
1.4		60.824033
1.5		63.474310
1.6		66.010599
1.7		68.442302
1.8		70.777704
1.9		73.024142
2.0		75.188147
2.1		77.275554
2.2		79.291602
2.3		81.241008
2.4		83.128038
2.5		84.956561
2.6		86.730094
2.7		88.451849
2.8		90.124765
2.9		91.751535
3.0		93.334637
Derivation of velocity and position vs. time for projectile subject to horizontal drag:

$$F_x = - \frac{1}{2}\rho A C_d v_x^2$$

$$m \frac{dv_x}{dt} = - \frac{1}{2}\rho A C_d v_x^2$$

$$v_x^{-2}\frac{dv_x}{dt} = - \frac{\rho A C_d}{2m}$$

$$\frac{d}{dt}(-v_x^{-1}) = - \frac{\rho A C_d}{2m}$$

$$(v_x^{-1}) = \int\frac{\rho A C_d}{2m} \, dt$$

$$v_x^{-1} = \frac{\rho A C_d}{2m} t + C$$

Initial conditions:

$$(v_x(0))^{-1} = C = (v_{0x})^{-1}$$

$$v_x^{-1} = \frac{\rho A C_d}{2m} t + (v_{0x})^{-1}$$

$$\frac{v_{0x}}{v_x} = v_{0x}\frac{\rho A C_d}{2m} t + 1$$

$$v_x(t) = \left[\frac{\rho A C_d}{2m} t + \frac{1}{v_{0x}}\right]^{-1}$$

Introduce constants to clear up the clutter

$$B = \frac{\rho A C_d}{2m}$$

$$D = \frac{1}{v_{0x}}$$

$$v_x(t) = \left[B t + D\right]^{-1}$$

Integrate this to find the position vs. time

$$x(t) = \int v_x(t)\,dt = \int \left[B t + D\right]^{-1} \,dt$$

Do a change of variables

$$u \equiv \left[B t + D\right]^{-1}$$

$$du = -B\left[B t + D\right]^{-2} dt$$

$$dt = -B^{-1}\left[B t + D\right]^{2} du$$

$$x = \int u (-B^{-1}\left[B t + D\right]^{2})\,du$$

$$x = -\frac{1}{B}\int u u^{-2}\,du$$

$$x = -\frac{1}{B}\int u^{-1}\,du = -\frac{1}{B}\ln u + C$$

$$x(t) = -\frac{1}{B}\ln [(Bt + D)^{-1}] + C$$

To figure out the integration constant C, evaluate this at t = 0, and set the position to be 0 there.

$$x(0) = -\frac{1}{B}\ln [(D)^{-1}] + C = 0$$

$$\frac{1}{B}\ln D + C = 0$$

$$\Rightarrow C = -\frac{1}{B}\ln D$$

Substitute this back into the original expression:

$$x(t) = \frac{1}{B}\left\{-\ln [(Bt + D)^{-1}] - \ln D\right\}$$

$$= \frac{1}{B}\left\{\ln [Bt + D] - \ln D\right\}$$

$$=\frac{2m}{\rho A C_d}\left\{\ln \left[\frac{\rho A C_d}{2m} t + (v_{0x})^{-1} \right] - \ln [(v_{0x})^{-1}]\right\}$$

$$=\frac{2m}{\rho A C_d}\left\{\ln \left[\frac{\rho A C_d}{2m} t + (v_{0x})^{-1} \right] + \ln v_{0x}\right\}$$

$$x(t) =\frac{2m}{\rho A C_d}\left[\ln \left(\frac{\rho A C_d}{2m} v_{0x}t + 1 \right) \right]$$
Attached Thumbnails

Recognitions:
Homework Help
 Quote by cepheid Cd is the drag coefficient
What value did you use for Cd?

Using the calculator linked to below, for a 6 mm sphere at 85 m/s, the Reynold number is a bit less than 40,000, at 40 m/s a bit less than 19,000.

http://www.grc.nasa.gov/WWW/K-12/airplane/reynolds.html