As a kid I tried a sling a bit. I had little idea how to use it and the only thing I ever hit with it was the ground.
Sadly, he uses both English and metric units some what willy-nilly. Wit hall due apologies to the counselor, 35m = 35 * 39.37" giving us 1377fps. Which is some suspect for being supersonic. Now, 125g (1929gr, circa 4oz) at 22-23m/s (8-900fps) equals rather a lot of joules., especially at 10-15m engagement range.
What does the "g" stand for in "500g" and "1000g?" I'm asking because there are 7000 grains in a pound.
Grams. Grains is not a metric unit of measurement. A kilogram (1000 grams) is 2.2 pounds, thus 500 grams = 1.1 pound.
So it's (mass x acceleration)? That's what this site says. Ah, well. It's been over 30 years since high school physics, so I don't feel too bad about missing that one. In any event, I still doubt that the force of a sling stone comes anywhere close to that of a .45, which I think is where this whole discussion started.
Spoiler alert - it’s momentum. You know, that pesky ballistic metric which is actually conserved during a real-world, inelastic collision - unlike Kinetic Energy which is NOT conserved, but sure sounds great when you’re trying to bamboozle unwitting fools who don’t realize bigger numbers don’t always mean bigger performance... It’s momentum... Not bothering to convert to traditional units, because the ratio remains the same either way: momentum P = m * v... A 1/4lb stone thrown at 35m/s (115fps) = .25lb * 115ft/sec = 28.75 lbm ft/sec A 230grn (0.03lb) bullet at 850fps = .03 * 850 = 27.9 lbm ft/sec. So a slung stone has greater momentum than a 45acp.
I wasn't trying to bamboozle anyone, so I hope that was a reference to the guy in the video. The running joke among lawyers around here is: I went to law school because I can't do math. Anyway, I guess dude was right.
Old testament Bible story I'm sure the stones used by David had fps provided by God to slay the giant
how about a sawn off M79 only 5 mm short of the goal If the sling and smooth flat river rocks are that powerful and have a five meter casualty radius I am going to have to start slinging again! Never noticed it before..... -kBob
Mea culpa, me maxima culpa. 35m/s = 35 * 39.37 = 1377.95 inches [yo estoy mui stupido] per second. or 114.8 fps, a much more moderate number. In my defence, I was flipping between conversion tabs and a KE calculator, and got caught up in the KE calculations. And then, I failed to confirm my wonky keyboard (for which a replacement is only 3 feet away) used "gr" for "grains" in the same sentence "g' was used for grams. Or exactly why the Mars Reconnaissance Lander tried to land in the Martian atmosphere at 13.2 m/s instead of 132 fps. (Serious geeky side note, the value in binary, 10000100, was the same, the problem was the call to the library of unit conversions was wrong; apparently the subr for the conversions were separated by 01 and 10, meaning no one considered being able to check for typos, visually.)
Neither energy nor momentum are equivalent to the force applied to the target. However the force applied to the target can be calculated using either energy or momentum. If one chooses to use momentum, then the force applied to the target is the momentum divided by the time it takes the target medium to stop the projectile. If one chooses to use energy, then the force applied to the target is the energy divided by the distance it takes the target medium to stop the projectile.
It was a reference to the wide-spread use of Kinetic Energy among the shootingsports world, which is, by and large, a meaningless metric on the business end, as collisions of bullets upon flesh and bone are inelastic, such kinetic energy is not conserved. Momentum, however, is conserved, and a far better measure of killing performance than KE. But people seem to like big numbers, and can relate to “foot-pounds” more easily than slug-ft/sec2, or lbm ft/sec.
Not correct. Momentum remains conserved in inelastic collisions. Kinetic Energy is not. Force applied to the target can absolutely be determined by the momentum transfer (assuming means to “know” the velocity of the bullet upon exit).
Still we do not know the weight of Little David's stone or his velocity.... but if he hit the big guy between the eyes with a .45ACP powered rock them cutting of his head was a redundancy.... When I got to using an arm length sling and rocks picked up from driveways, planters, and such good enough to hit a coffee can at 15 yards maybe one and three times I never did more than dent the can....do I need to 'spain what cans like that I shot with a .45 ACP even at 100 yards looked like? No doubt it I had actually lucked out an smacked Big G on the bridge of the nose he may well have been hunched over holding his face with both hands while I did a modified Tueller Drill Run that ended with me picking up his sword and using it on him. On the other hand a shot to the same place with my service issued 1911A1 with a 230 grain FMJ round would have resulted in a hole in the back of his head and perhaps his bronze helmet as well. Other than that 45mm gaff and I believe the power beliefs I liked his talk and his thoughts on the importance of Pelltasts in early war. BTW with a single cord tied on a Mark II or M 26 frag grenade and a buddy or fixed C rat can to hold the spoon down and a three quarter loop release you can easily throw the grenade far enough that it is still in the air when it explodes sort of like milking it and cooking it off but with out the short fuze danger and easily near twice the distance you can throw it from your hand. The classic hammer throw is but a spoterization of this very style of delivery and was used with large rocks (AKA BFRs) fore instance to break up a tight spear and shield or sword and shield formation. Some say they were even used against the earliest Brown Bess Armed English troops in Scotland. Slings were neat and if my shoulders were not shot would still be fun but I am reminded of a drunken discussion with the late Robert Asprin (aka Yang T. Nauseous) where he maintained that he could take me on with his English Long bow at 600 yards and I with my M 1903 A3 and ball ammo.... I still feel that he would have lost that one -kBob
Kinetic energy and Momentum have a lot in common. They should have given that they're both made up of the same two ingredients (mass and velocity). Conserved or not, any way you cut it, the application of force generates kinetic energy and the application of force reduces kinetic energy. In exactly the same manner, the application of force generates momentum and the application of force reduces momentum. One can determine the force required to generate a given amount of kinetic energy by measuring the distance it took to generate the energy, and therefore, one can also determine the force generated by dissipating a given amount of kinetic energy by measuring the distance it takes to dissipate it and dividing the energy change by the distance. In other words, change in energy with respect to distance is the same thing as force. In exactly the same manner, one can determine the force required to generate a given amount of momentum by measuring the time it took to generate the momentum, and one can determine the force generated by dissipating a given amount of momentum by measuring the time it takes to stop the moving object and dividing the momentum change by the time. In other words, change in momentum with respect to time is the same thing as force. This isn't something I came up with on my own, although it's not difficult to see the relationship by going through basic unit analysis on force, momentum, acceleration, energy. https://physics.stackexchange.com/questions/370960/relationship-between-net-force-and-kinetic-energy https://sciencing.com/calculate-force-impact-7617983.html "Therefore, work is equal to force multiplied by distance: W = F × d. Because force is a component of work and an impact is the conversion of energy into work, you can use the equations for energy and work to solve for the force of an impact." https://courses.lumenlearning.com/physics/chapter/7-2-kinetic-energy-and-the-work-energy-theorem/ "(a) Calculate the force needed to bring a 950-kg car to rest from a speed of 90.0 km/h in a distance of 120 m (a fairly typical distance for a non-panic stop). (b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force exerted on the car and compare it with the force found in part (a)." https://www.gigacalculator.com/calculators/impact-force-calculator.php Impact force formula The formula for impact force expressed in terms of the body's velocity (speed) on impact (v), its mass (m), and the collision distance (d) is the first formula below: whereas the second one is applicable if instead of the collision distance we know the duration of the collision (t) Note that the first formula is kinetic energy divided by distance and the second is momentum divided by time.
My post quoted gr, not g... 500gr - grains = 30g approximately... In grams, they find lead bullets from 30 to 60 grams, with 50 grams seeming standard in many places.
1m = 3.28', you used inches... 35 m/s = 115 ft/s But tests done with skilled modern slingers got higher velocities. Trained army slingers, 2000 years ago, would have danced circles around modern guys slinging for fun and to kill rabbits. Just like the English longbows were 100lbs and more... Who can pull a 100lbs bow nowadays? For sling performances comparison, tennis balls weigh around 2oz, 850gr to 900gr. The fastest tennis ball was clocked at some 240fps. That's a projectile launched with only the strength of the sportsman's arm, and an extension. When thrown by a highly trained man, it's not far fetched to expect velocities around 300fps to be attained with a lighter projectile offering better aerodynamics, as a lead sling shot is...
Now, to keep it simple... Try an ounce and a half of lead, hitting you in the face at 150mph or more... Do the figures hit home now? Like kBob says, cutting Goliath's head off was mostly for show...
As guyfromohio already stated, this is a rubber slingshot and not the leather-strap type of sling used in ancient times. They're totally different in how they work, but the type, mass and velocity of projectile they use are similar to one another in contrast to handguns. I suppose it's like the ultimate showdown of "slow and heavy" vs. "fast and light." If you're wondering about the performance in ballistic gel, there are videos on that too. Even with a pocket slingshot, the penetration can be very substantial. I have not seen slung shot that expanded, but much of it starts out with a much larger diameter than most handgun bullets do. Where the handgun is certainly going to maintain an big advantage over slings of both types is in range. The handgun can easily be very deadly at 100 yards, whereas it's doubtful a sling could achieve that range or be effective at those distances.
Nope. Kano said the Romans used projectiles from 500gr to 1000gr. The gr stands for grains and it takes 7000gr to make 1 pound
147 HST+P @ 1050 ft/s = 22 lb-ft/s 50 gr TSX @ 3090 ft/s = 22 lb-ft/s Hmm. And I think the best example is that of firing a gun. Momentum is equal between the two; both the recoiling gun and the flying projectile have the same momentum. Kinetic energy is not equal... and because of this, being hit by the bullet is a lot worse than being hit by the gun.
@JohnKSa: what your copy paste work above is failing to recognize - AGAIN.... In your “first equation,” Favg = .5 m v^2 / d, recognize the m and v are NOT the mass and velocity of the bullet, they are the mass and resulting velocity of the affected mass. Because kinetic energy is NOT conserved, we do not know what the velocity would be. We might know the mass of the target - which largely is not true, since we don’t get to know the mass affected by the Work done in a gunshot wound, but we can pretend for a moment. We don’t have a means of quantifying the Work done, or the force involved, because we know kinetic energy is not conserved in an inelastic collision. Considering the gunshot wound system, we’d have to determine some approximation of temporary cavity mass and instantaneous velocity, less the coefficient of elasticity of the tissue which exerted a resisting force against the work being done, then a differential velocity exposed to a differential mass until we reach a net velocity for the gross mass of the target. This is an exceptionally complex system to solve, with enough assumptions forced into it to completely invalidate the output. Even if a bullet stops within the target, we cannot be sure the total kinetic energy on the bullet at impact was transferred (KE:W = 1) as kinetic energy. Deformation of tissue and of the bullet causes LOST energy, even the sound of the “smack” of the bullet on the target or the heat created from the friction of the contact or the heat of the compression of the tissue, and fragmentation of the metal bullet are all means of kinetic energy on the bullet which is NOT transferred to kinetic energy on the target. All of these losses confirm we KNOW all of the kinetic energy on the bullet is not transferred to the target. Energy can be converted into many forms. HOWEVER.... In the second equation, Favg = m v / t, we DO have a simple means to know the mass and velocity terms. Momentum is conserved, such if we know if the bullet stops in the target, or we can determine the velocity of the bullet upon exit, then we know the momentum transferred to the target. Here’s an example. A 100grn bullet fired at 1000fps into a 900grn block, in which it stops. Not messing with unit corrections here to keep life simple. *Reminding before we start: KE = 1/2 m * v^2, but in the context of ballistics, we convert grains to pounds and have to convert weight to mass, such it becomes KE (ft.lbs) = (bullet weight in grains) * (velocity in ft/s)^2 / (2 * 32.17 * 7000). Such kinetic energy formula, with unit conversion, is: KE = m* v^2 / 450380. The block is at rest to start, such it has no momentum to contribute to the initial state. Momentum bullet upon firing 100 * 1000 = 100,000 grn ft/s Kinetic energy of bullet upon firing 100 * 1000^2 / 450380 = 222 ft.lbs. Upon collision, the bullet sticks (stealing some graphics from online too): The net mass now is 1000grn: bullet + block. The net velocity follows: 100 * 1000 = (100 + 900) * V2. Pretty easy math here, the resulting velocity will be 100ft/s. So the net momentum of the post-collision block is (100+900) * 100 = 100,000 grn ft/s. So let’s see how your Kinetic Energy transfer faired... KE = (100+900) * 100^2 /450380 = 22.2ft.lbs. Momentum is conserved, kinetic energy is not. Blathering on about kinetic energy in this discussion doesn’t make sense. The science is outlined above. In this example, as always, 100% of momentum is conserved, while only 9% of the available kinetic energy was actually transferred to the target (remaining 1% on the bullet as they travel together).