.55 cal paintballs ... new FPS limits?

[slade]

Exactly.... If you get a bowling ball traveling at the same speed as a baseball, the bowling ball will travel further than the baseball.

Pneumagger makes some good points also.

1) notice i said it was a bad analogy
2) pneumagger may have made some good points, but you didnt.
3) no, it WONT necessarily go farther! thats the whole damn point! please explain your (lack of) logic here.

any two bodies, regardless of mass, which have the same velocity in an enviroment with no air resistance will travel the same distance. its not that hard of a concept. therefore, the distance the object travels will be dependent on air resistance and how it effects the momentum of the object. the only thing you can conclude from that is if there are two objects with the same surface and size, but one is less dense than the other, the more dense one will travel farther. AKA, if you have a paintball with a fill and a paintball that is empty traveling with the same velocity, the full paintball will travel farther.

therefore in determining the distance an object will travel, you not only have to take into account mass but also surface area, shape, surface qualities, spin, etc. your thoughts are far too simple and misguided.

[bjjb99]

This is a Deep Blue post, fair warning...

Let's look at two cases... The first case will be the effect of different velocity "headwinds" experienced by a .55" paintball moving at 350 fps versus a .68" paintball moving at 300 fps. The second case will examine the effect of a constant velocity crosswind on each of these paintballs.

Aerodynamic drag is given by the following formula:

Fd = 0.5 * Cd * rho * A * V^2

where
Fd = drag force in newtons
Cd = drag coefficient
rho = density of the fluid (in this case, air 1.29 kg/m^3)
A = cross sectional area of the object in m^2
V = velocity in m/s

Consider the drag coefficient. It is a function of the Reynolds Number for the object moving through a fluid. If you compute the Reynolds numbers for the 0.55" paintball at 350 fps and the 0.68" paintball at 300 fps, you find that the Reynolds numbers are within 10% of each other. Mapping Reynolds numbers to drag coefficients for spheres yields a value of right around 0.4 for both the 0.55" and the 0.68" paintball.

The area of the 0.55" paintball is 0.000153 m^2 and its velocity is 106.7 m/s.
The area of the 0.68" paintball is 0.000234 m^2 and its velocity is 91.44 m/s.

Now that we have all the numbers, we can calculate the drag force on each paintball at their respective exit velocities:

Fd(0.55) = 0.5 * 0.4 * 1.29 * 0.000153 * 106.7^2 = 0.45 Newtons
Fd(0.68) = 0.5 * 0.4 * 1.29 * 0.000234 * 91.44^2 = 0.50 Newtons

These values agree well with results obtained from a Javascript drag calculator located at the following URL:

http://www.fluidmech.net/jscalc/cdre01.htm

The forces acting on the two paintballs are only different by 10 percent. However, their effects are quite different. The acceleration of each paintball is determined by the force acting on it and its mass. If we assume that the 0.68" paintball has a mass of around 3.10 grams (a reasonable average) and if the 0.55" paintball has the same fill and shell composition, then the 0.55" paintball will have a mass of around 1.64 grams. The respective accelerations will be as follows:

a = F / m
a(0.55) = 0.45 Newtons / 0.00164 kg = 274 m/s^2
a(0.68) = 0.50 Newtons / 0.00310 kg = 161 m/s^2

The 0.68" paintball will decelerate much slower than the 0.55" paintball. The net result is that while the 0.55" paintball has a higher exit velocity, it won't travel as far as the 0.68" paintball with a lower exit velocity.

[edit: To first order, the 0.68" paintball travels about 8 meters farther than the 0.55" paintball for shots fired horizontally at 5 feet above the ground]


Now let's look at a crosswind and how it affects each paintball. Let's assume a very slight wind, about what you'd feel on your face if you were just walking through still air (6 feet per second crosswind). I'm going to just use the Javascript drag calculator I mentioned above rather than go through all the computations yet again.

The crosswind force on the 0.55" paintball in a 6 fps wind is 0.0016 Newtons, while the crosswind force on the 0.68" paintball in the same wind is 0.0024 Newtons -- a full 50% higher! Take a look at the accelerations, though:

a(0.55) = 0.0016 Newtons / 0.00164 kg = 0.98 m/s^2
a(0.68) = 0.0024 Newtons / 0.00310 kg = 0.77 m/s^2

The 0.68" paintball is blown off target less by the crosswind. In higher winds, the 0.68" paintball wins out even more.

Based on the above analysis and my own personal observations of 0.68", 0.62" and 0.50" paintballs back in the late 1980s, I'd have to say that the 0.68" paintball travels farther and with less wind deflection than smaller, lighter, faster paintballs.

BJJB

edit -- added first order range difference

[Pneumagger]

1) notice i said it was a bad analogy
2) pneumagger may have made some good points, but you didnt.
3) no, it WONT necessarily go farther! thats the whole damn point! please explain your (lack of) logic here.

any two bodies, regardless of mass, which have the same velocity in an enviroment with no air resistance will travel the same distance. its not that hard of a concept. therefore, the distance the object travels will be dependent on air resistance and how it effects the momentum of the object. the only thing you can conclude from that is if there are two objects with the same surface and size, but one is less dense than the other, the more dense one will travel farther. AKA, if you have a paintball with a fill and a paintball that is empty traveling with the same velocity, the full paintball will travel farther.

therefore in determining the distance an object will travel, you not only have to take into account mass but also surface area, shape, surface qualities, spin, etc. your thoughts are far too simple and misguided.

Yup. I know. (I'm an aerospace engineer)
I was trying to keep the post as simple as possible. But suface area is a large factor too. I was trying to make the point that drag forces (from crosswind) can effect a lesser mass more easily.

But yes...there will be less surface are on the ball. I just assumed that a slightly lesser area coupled with a significantly faster speed (as stated previously by someone) would result in at least the same drag or more on the smaller ball.

Also, wouldn't the smaller radius lend itself to more frequent shedding vortecies?

[Dend78]

^^ yes im an avid firearm shooter i dont know all the formulas but i do know that the lighter the projectile becomes the more eaisly wind effects it due to there not being the weight to keep it stable, but i dont know since its still a round ball its not gonna cut the air any better so i would assume about the same distance, possibly a little farther but nothing extravagant like 100yrd flat shots,


but that brings a question to mind i know that the miltary and LE use paint rounds for rifles and things for various training exercises, these are cone shaped like regular bullets from what i understand, will rifling actually work on these seeing as you still have to try to spin the paint as fast as the shell, i would actually like to examin one of those rounds up close, just to see or get an idea how it works, like i said i dont know my formulas and stuff but it still makes me wonder how those rounds work, i know the twist rates and the pressure would be much higher out of a rifle rather than a paintball gun but still getting the paint to spin at the rate of the shell would be very tough i would think

[personman]

Too easy to wipe.