drag coefficient g1
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drag coefficient g1 Q&A Review
How would you calculate a bullet's velocity reduction due to drag?
It can be estimated reasonably well by using a drag coefficient. The formula for drag force is F = C*density*A*v^2/2 where A is the cross section area (pi*r^2), v is the velocity, density is mass/volume and C is the coefficient (usually written C with subscript d for drag). The value of C changes with bullet shape and velocity, but you can find lots of stuff on the web that estimates it, like: Since the value of C changes with velocity, the simplest thing to do is a numeric analysis using Excel. Have a table with the top line being, for example of starting at Mach 2, which is 2450 ft/s: Time (=0); Velocity (=2450 ft/s); C=0.6; F=C*dens*A*v^2/2; a =m/F Then in the second row, make the time increment by some small value, like 0.001 seconds. The formulas become: t=prev row+.001; velocity=prev v - a*.001; C=lookup; F=C*dens*A*v^2/2; a =m/F Then repeat for 1000 rows and you get a velocity history for 1 second of flight. The lookup feature would involve you manually writing a table of values of velocity and C, like a column of velocities in increasing order from 0 to mach 5 (6125 ft/s) in, say, cells F1:F30. Then manually add C values in cells G1:G30. I just picked 30 rows, you can add the data as you wish, just use plenty of points around where the velocity changes the most. Assuming your velocity is in cell b5, the lookup formula is =lookup(b5,$F$1:$f$30, $G$1:$G$30). That formula uses the $ to fix the cell, so that when you copy the formula down a column, the b5 number changes to b6, b7, b8, etc., but the reference table doesn’t change.
A bullet is shot vertically upwards with an initial velocity of 50m\s. How high does it travel?
THAT will depend on three things: the mass (weight) of the projectile; its cross sectional density; and it’s profile , which in turn will determine its ballistic coefficient. Mass and profile (shape) are static. Contrary to common belief, ballistic coefficient is depend on velocity at any particular point in time. Since 50m/s is a subsonic speed, there will never be any transonic effects. Ballistic coefficient is a measure of drag vs. mass. A high BC means low drag, hence the bullet will travel further with the same initial velocity. There are a number of “standard” models (standard bullet shapes) from G1, roughly like the round nose of a .22 with a straight can shape, to the G7, with fairly pointy nose and a boat tail (beveled tail). G7 bullets are used by snipers, hunters and other long range target shooters - these projectiles have ver high BCs. it is important to note that BCs change, dependent upon the velocity, and the air density. A bullet fired straight up -assuming no effect from wind- will experience LESS drag as it climbs, but since in your example 50m/s is a relatively LOW velocity, the air density change will be minimal. Here are the formulae for calculating maximum height: h = v0 *th*sin(90) - 1/2*g*th^2 h = (v0^2*sin^2(90))/(2g) where v0 = initial velocity; th = time to height; g = gravity constant.
Can you be shot from so far away that the bullet loses enough energy and the shot does not break skin and just hurts?
Can you be shot from so far away that the bullet loses enough energy and the shot does not break skin and just hurts? There are lots of anecdotal cases of this happening, but then there are a very wide range of guns, bullets and circumstances to consider. To have a sensible discussion about the mechanics we need to narrow this down. Let’s go with a 9mm pistol together with a bullet that leaves the muzzle travelling around 350m/s and having a G1 ballistic coefficient (bc) of 0.15. In case you’re not familiar with bullet trajectory modelling, the bc is a measure of how well the bullet maintains its speed despite air resistance, aka drag. G1 is what’s called a reference projectile for which there are tables available of deceleration due to drag against projectile speed. You divide this reference by the bc for the bullet you’re considering to get the deceleration for this bullet for a given speed. A bc of 0.15 indicates quite a slow bullet meant for short range, but this will give us a reasonable chance of getting slow enough speeds at longer ranges. We also have to decide how fast a bullet needs to go to break the skin. Again this is very variable but if we consider the bullet striking head on then a generally accepted rule of thumb is 60m/s. Obviously bullets can travel much faster and not break the skin if they strike at an acute angle and just get deflected. We’ll also fire at the same height as the target, which I’ll take as the zero height reference, and arrange for the target to be at the right distance for the trajectory to hit at this height. It’s a calm day with no wind or air current to take into account. To get the bullet to travel a reasonable distance then we need to fire at an angle to horizontal. Let’s start with 10˚ and see what happens: Here the blue curve shows the trajectory with the height scale on the right, and the red curve shows the bullet speed with the speed scale on the left. In this case by firing upwards a little we’ve managed to get the bullet to travel 1300m before hitting the ground, and sure enough the speed has dropped considerably on the way, to around 100m/s. But this is still around 2.8 times the energy compared to our target 60m/s. So let’s try turning the angle right up to 50˚ (all charts use the same scales for easy comparison): Well, the range has now gone out to 1550m, but the speed on impact is still 90m/s. And you can see why - the bullet has gone high enough to gain a lot of speed from gravity on the way back down, despite going below 60m/s at the top of the flight. So instead we have to make a compromise, giving the bullet enough height to travel long enough to slow down but not enough for it to speed up again too much on the way down. It turns out that with this particular gun and bullet the best we can do is firing at around 23˚: Even then the speed on striking the target is 79m/s. This is still significantly too fast, having around 70% more energy than our target 60m/s. So no, we can’t make the bullet go slow enough in this scenario. We’ll have to do something different if we want a slower impact speed. We could find a bullet with a slower muzzle speed, but it turns out that lowering the muzzle speed doesn’t actually have much effect at this distance. Slower starts just mean that the bullet doesn’t decelerate due to drag as much early on in the flight. The range reduces but not the impact speed, illustrated here with a 250m/s start: How about introducing a headwind instead, say a steady but brisk 20m/s? Firing at 23˚ 350m/s the headwind has reduced the range again but has only decreased the impact speed to 68m/s. Unrealistically high winds are required to get us down to the target impact speed. Of course we can always raise the height of the target relative to the firing position, but that’s definitely cheating! The bottom line is that with an ordinary pistol we can’t get down to our target speed of 60m/s which is needed to have a good chance of not breaking the skin with a head-on strike. Our best bet is to fire into the wind at around 23˚ above horizontal and have the target at an angle to the bullet to reduce the impact. In this situation we would probably manage not to break the skin and so the answer to the question would be yes, but only by increasing the odds in your favour.
How hard is it to hit a human sized target at 2000 meters with a rifle? What factors need to be taken into account?
I’ll give you a hint; only 8 soldiers, out of the hundreds of millions who have served since the invention of the rifled firearm, have recorded a confirmed kill at or beyond 2000 meters with any small arm. Chris Kyle missed out on being the 9th such sniper by just 80 meters. Wikipedia editors start paying attention at 1250m, and only 19 snipers in history have recorded a confirmed kill longer than that. (Edit: August 2018 saw a 20th shot, Carlos Hathcock style: ,SAS hero sniper takes out ISIS commander with just one bullet from over a MILE AWAY,. Exact distance isn't officially known yet) So, here’s how you line up a 2000-meter shot. It starts with: BALLISTIC ARC (BULLET DROP) The primary determinants of the path of a projectile moving in a gravitational field are the initial velocity of the projectile in the X (horizontal aka parallel to level ground) and Y (vertical aka parallel to gravity) axes, the acceleration of gravity (9.8 m/s^2), and the deceleration of drag, all occurring over the time of flight. The calculation of the velocity, position and time of flight of a bullet moving through air at high speed are college-level differential equations, because the drag on the bullet both affects and is affected by the bullet's velocity, so the equation of drag force over time given an initial velocity (and thus the related equation for velocity over time accounting for drag, and its integral for position over time) is most elegantly defined as a function of its own derivative. This is why your high school physics professor kept telling you to neglect or ignore drag in projectile motion calculations for that class; taking it into consideration makes everything ,much, harder. Luckily for non-engineers who likely didn't see this math even in college, the calculation can be simplified to an algebraic formula for range or time of flight using the caliber, mass, muzzle velocity and “ballistic coefficient” of the bullet, and the density of the air (a function of temperature, barometric pressure and humidity): x = vt - (CρAv^2t^2)/2m where ,x, is horizontal range, ,v, is muzzle velocity, ,t, is flight time (which you have to rearrange to solve for), ,m, is mass, ,C, is the ballistic coefficient of the bullet (the relative aerodynamic “efficiency” of the bullet compared to an ideal model; ammunition manufacturers will provide this number), ,ρ ,is air density and ,A, is the cross-sectional area of the bullet (pi*(diameter/2)^2). This particular model simply assumes a linear deceleration from the muzzle velocity, which is close enough to get civilian precision shooters in the neighborhood and empirically adjust from there. You can rearrange the above to solve for t given a known x (hint; remember your logarithm properties), or just plug these values into a spreadsheet formula and goal-seek. However, for 2000m, linear deceleration produces enough error to be ,way, off. Here’s a more complex model: C_d=\dfrac{.5191m}{C_bd^2} C_{Δv}=\dfrac{500}{m}*\dfrac{1.225C_d\pi(\dfrac{d}{1000})^2}{4} t=\dfrac{e^{rC_{Δv}} - 1}{C_{Δv}v_x} This is a more accurate calculation based on the differential equation, which when given accurate values for the following, tends to work pretty well: C_b, = ballistic coefficient of bullet (G1 method) C_d, = drag coefficient (the above equation hardcodes atmospheric conditions, namely fluid density; the constant .5191 really isn’t constant) C_{Δv}, = coefficient of deceleration (basically a useful substitute for the rather complex equation on the right-hand side of its definition) m, = bullet mass in grams d, = bullet diameter in millimeters r, = range in meters v_x, = muzzle velocity along the ground in meters per second (muzzle velocity along the bore times cosine of launch angle; can be used to compensate for sight-in range and resulting upward bore angle with the scope “level”) t, = time of flight in seconds However you calculate time of flight, you then plug it into the high-school physics equation for distance over time given acceleration of gravity - ,d = .5gt^2, - to calculate bullet drop from boresight. This calculation neglects drag entirely, and that’s also acceptable as the amount of drag in the vertical from gravity really is negligible relative to the frontal drag, even over this distance. Now that you have the range and bullet drop, you simply compensate for it in the scope by dividing one by the other; if you’re using yards of range and inches of drop, dividing bullet drop by range and multiplying by 100 gives you minutes of angle or MOA. If you’re using meters and centimeters, dividing drop by range and multiplying by 10 gives you milliradians or MRAD (sometimes called MIL, but while mathematically there are 2000π ~= 6283 MRAD in a full circle, other values for MIL have been used including 6000, 6300 and 6400, and at 2000 yards it makes enough difference). The units you use will dictate the type of scope you get. You then compensate for the angle between boresight and optic; the scope is above the bore, and is “looking down” at where the bullet will be at the distance you sighted the rifle in. So the scope height and sighting in distance gives you another MOA or MRAD angle which you can simply subtract from the bullet drop compensation. WINDAGE Then, you just need windage. Assuming a constant wind speed and direction relative to your bullet’s flight path, the wind will produce a constant force and thus a constant acceleration of the bullet perpendicular to travel, much like gravity will create a constant bullet drop. The equation for wind pressure given wind speed is basically the drag force equation: ,F_d = .5Cρv^2,, where C is the drag coefficient of the bullet’s side profile. For simplicity, use 0.4 (drag coefficient of a frictionless perfect sphere; our bullet is neither of these but it’s close enough). Velocity is meters per second, so ,F_d, will work out to newtons per square meter aka pascals, making this a pressure calculation instead of force. You then need the cross-sectional area of the bullet in profile to get the actual force in newtons, which with the mass of the bullet gives you acceleration in the direction of the wind, which then gives you deviation over the time of flight, that you can convert to an angle and correct for in the optics. Oh, and don’t forget that if the wind’s not exactly perpendicular to the bullet’s path of travel, a component of it will be helping or hindering the bullet by changing the speed of the relative wind, pretty much as if the muzzle velocity were faster or slower by the amount of wind blowing in line with the barrel. The cosine of the relative angle of wind, assuming your bore is “zero”, gives you the component of wind speed required to adjust muzzle velocity. Then you have to go back and do the ballistic arc calculations again with that new muzzle velocity. GYROSCOPIC DRIFT/SPIN DRIFT Now, in calculating bullet drop, we ignored drag in the vertical, which is usually negligible unless you’re firing with enough elevation angle from level that the bullet approaches terminal velocity on the way down from apogee; if you’re within 20 degrees of level with the ground, it probably won’t get there. However, over long enough distances, the force of air drag in the vertical can induce gyroscopic precession, also known as “spin drift”, which also has to be accounted for in a 2000m shot. Simply put, the spin imparted to the bullet by the rifling will stabilize the bullet around its long axis as it flies though the air, keeping that axis pointing parallel to the rifle bore. So, as the bullet begins to descend from the peak of its arc, it will still be pointing upward relative to its direction of travel, and so the oncoming air will be pushing on the nose of the bullet from underneath. When you push on a spinning gyroscope’s axis, what you get is a rotation of that axis at a 90 degree angle to the direction of the force applied, based on the direction the gyroscope is spinning. So, with most rifles having a right-hand twist, as the bullet falls through the air, the nose of the bullet will be pushed to the right, and like a plane’s rudder in the wind, as the bullet turns it presents its side to the oncoming air, which will cause the bullet to veer right. The calculations for spin drift depend on twist rate and muzzle velocity, which gets you the spin rate in RPM, then the mass and diameter of the bullet give you the moment of inertia, and the length of the bullet gives you the torque arm length from center of mass. You then find the force applied to that torque arm by calculating the component of the force of the relative wind that is perpendicular to the bullet’s spin axis. The force of the relative wind is the combination of the instantaneous drag force caused by the bullet moving and the drag induced by the bullet falling, and the ratio of these gives you a compound force and angle of relative wind relative to level. You add the bore angle at launch (which the bullet is still spinning along, at least vertically), compute the perpendicular component of the force of this wind, and then apply that to gyroscopic precession equations given the mass, spin rate and moment of inertia of the bullet to determine the precession amount. Then, a split second later, that precession amount changes everything, making this another differential equation most elegantly defined in terms of its own derivative function. CORIOLIS EFFECT/EOTVOS EFFECT The Coriolis Effect, a major subcomponent of which is the Eötvös Effect, is another element to consider, and it’s a weird one because it depends not only on which direction you’re shooting, but where you are on the planet. Broadly speaking, bullets that actually make it 2000m have to go fast enough that they are for our purposes suborbital, requiring you to consider the effects of the bullet’s travel relative to the movement of the Earth itself. Generally speaking, the Eötvös Effect means that if you shoot east, the bullet’s velocity combines with the Earth’s rotational velocity to increase centripetal force of the bullet in the rotating inertial frame of the spherical surface of our planet, causing the bullet to fly higher. Shoot west, and exactly the opposite happens; Earth’s rotational speed in the other direction cancels out some of the velocity for these purposes, and the bullet’s centripetal force against gravity will be lessened, causing it to drop more over the same range. This effect is more pronounced the closer the compass direction of your shot is to east or west, and the closer you yourself are to the Equator. Shooting north or south, there’s an additional lateral component to the effect more directly attributable to Coriolis; shooting away from the equator, your bullet will be travelling faster to the east than the ground it moves over, causing the bullet to impact to the east of the target by enough over 2000m to miss by. Shooting toward the equator, again the opposite happens; the ground is moving faster east than the bullet is, so the bullet will impact to the west of the target. By how much? Depends on the ratio between the distance inherent in a minute of longitude at the shooter’s location versus the target’s, which means the closer you are to one of the two poles, the more it matters. So, on the northern hemisphere, shooting north will cause the bullet to land to the right, shooting east will bring it high, shooting south will lead it left, and shooting west will drop it low. On the southern hemisphere, reverse the north-south behavior, but the east-west behavior is identical. The east-west differences are more pronounced the closer you are to the equator; the north-south distances matter more the closer you are to the poles. Confused yet? SUBTLETIES OF RANGING Now, I’ve assumed so far that you just happen to know the ,exact, range of your target. In the real world, even in the relatively sterile and static environment of a shooting range, targets are rarely ,exactly, at the stated distance relative to your muzzle. There are many tools you can use to obtain this range; few are totally exact over this kind of distance. The most accurate is a laser rangefinder, similar to what golfers use, but golfers typically only need to know ranges out to about 500 yards; you want to go four times further out. Most of the civilian rangefinders aren’t guaranteed to be accurate longer than 500 yards, and aren’t guaranteed to give you a reading at all beyond 1000. There are a few 3000-meter rangefinders out there, starting at $500 with the ones I’d actually trust at that distance costing more like three times that. Don’t have $1500 for a rangefinder? Fear not; you can use your scope’s subtensions and more math. If you’re making a 2000-yard shot, you’re using a calibrated, graduated scope that will allow you to measure the angular size of your target in MOA or MRAD. If you know, or can guess accurately enough, the actual size of that target, you can determine the range to target using much the same trigonometry as the bullet drop compensation. Let’s say you get a good, front-on look at your human size target, and you measure the width of his shoulders at .25 MRAD. The average man’s shoulders are 46cm or .46m wide. Assuming your target is of average build, this will be close enough. The MRAD ranging formula is just a rearranging of the bullet drop compensation formula; actual size in meters, divided by measured MRADs, times a thousand, is the range in meters. For this example, your target is 1840 meters out. Keep in mind, when doing this range estimation, that this is what a typical graduated MRAD first focal plane scope looks like at 8x and at 34x magnification: Just for context, your target’s shoulders, at the stated range, would be the width of one ,side, of the hash marks on the vertical crosshair (from the vertical line to the edge of the hash mark). If he’s 1.8m tall (5′11″), standing straight at this range, he’s just a hair smaller than 1 MRAD total height (from the crosshair up to the first full-length hash). And at this distance, if you misjudge range by even 10 meters, the difference in bullet drop between 1840 and 1850 meters is enough to plant your shot in the ground at his feet, or shoot over his head. Not rocket science, but pretty close, actually. Luckily for the amateur precision shooter, all this math has been programmed into a number of ballistic calculators, such as Strelok and the Lapua Ballistics app for Android (if you’re an iPhone type, you’ll have to find your own). You’ll need the caliber, mass, muzzle velocity and ballistic coefficient of the bullet, then the range to target, temperature, barometric pressure, relative humidity, wind direction relative to the barrel and average wind speed. The apps will make some assumptions about the rest of the variables like profile cross-section, and will get you pretty close. Most calculators I’m aware of don’t figure for spin drift, Coriolis effect, etc, but there are a few. The mile shot, 1760yd or about 1610 meters, is not an impossible shot if you know what you’re doing, and you have enough chances. Here’s a YouTuber going for a little more than that, 1820 yards (1664m), with a relatively inexpensive rig based on a Ruger American Predator rifle chambered in 6.5mm Creedmoor. With scope, rings and bipod, about $750 total investment at the time, though the rifle’s gotten a little more expensive since then. For context, the Arctic Warfare Magnum in .338 Lapua, which is a common high-end precision shooting rifle and holds the #2 longest-distance confirmed kill as the L115A3 in British service, streets for $8500, and then you’d put about a $3000 scope on it. So this guy’s going for the mile with a rig less than one-tenth the cost of a purpose-designed precision rifle. Anyway, the shooting ,(EDIT: my apologies, but it looks like this YTer’s account has been closed due to YT’s new policy regarding firearm-related videos; below is the same video on DailyMotion, which doesn’t embed into Quora),: Shooting a Mile with Ruger American Predator!? (1820 yards) - Video Dailymotion Now, that’s 4 hits out of 12 shots. After his first shot based on precalculated values for bullet drop given the ammunition he was using, his spotters recommended a full MRAD up and another MRAD to the left, so he needed to get about 1.4 MRAD closer to center (which at this distance was 2.33m or 2.55 yards off) just to get in the neighborhood. He then needed another .6 MRAD up (.998m, 1.1yd) and another .75 MRAD left windage correction (1.25m, 1.37yd) over the next two shots, before his fourth shot made the first hit. So his first shot, the one that, for a sniper, is usually the only shot, was over 2.66m low and almost 3m to the right, almost 4m off the intended point of impact all together. That’s not just missing, that’s not even close. After his barrel warmed up, the shot pattern started to rise and to spread out as the barrel became more flexible. So, you don’t get too many accurate mile shots with one rifle in a particular time frame. What was off? Any of a number of things: Neglecting the actual air density in the precalculation of ballistic arc, which can’t be known ahead of time and is commonly based on average Earth temperature and humidity in the temperate zones of 60*F (15*C) and 45% relative humidity at sea level (~1000mbar). Cooler, less humid or higher-pressure air than that will be denser, increasing drag, time of flight and bullet drop. No initial correction for wind at all; even knowing the wind speed at the shooter’s location ,and, the target’s, over a mile of distance the wind speed and direction can vary enough that averaging these two known data points is wrong enough to miss by. No apparent correction for spin drift; the 6.5 Creedmoor is a very long bullet for its diameter and mass, which increases ballistic coefficient but also the amount of gyroscopic drift, as the oncoming air has a longer lever arm to push against. Also no apparent correction for Coriolis Effect; this shot’s long enough for it to matter. Given he was on the northern hemisphere in the U.S. for this video, for it to be adding to the error we saw in the first shots, he would have been shooting to the northwest. All these factors would have to have been considered and accounted for correctly in order for him to have made the target on his first shot. That’s in addition to simple bullet drop due to gravity which he did account for, but most likely only based on average performance of the 6.5 Creedmoor round in average or ideal conditions; he would have had to adjust for any difference in temperature, humidity, altitude etc relative to the conditions for which the bullet drop calculations were made. And this distance is 336m shy of 2000m. That’s ,how hard hitting a human-size target at 2000m with a rifle is.
What is the ideal ammunition grain weight for a rifle with a 1:9 twist rate?
Ideal is more than the weight of the bullet. Design and materials factor into it. The classic method of determining whether a bullet will be stable at a given twist rate is the ,Greenhill Rifling Formula, ,developed in 1879 by Sir A. G. Greenhill. Below is the common implementation of it: Where D is the bullet diameter, L is the bullet length, SG is the specific gravity of the bullet, and C is a constant (150 for MV<=2800fps and 180 if greater). All measurements are in inches. Common values for SG are: Lead 11.3 Jacketed 10.9 Copper 8.9 Brass 8.5 Steel 7.8 This formula has been improved on since then. Don Miller improved on it , ,. Adaptations of this formula have been made to allow inputting the ballistic coefficient and unfixing variables he fixed for simplicity, such as temperature and altitutude. Several web sites have automated the formula. One of the most comprehensive is from Berger Bullets. Twist Rate Stability Calculator | Berger Bullets This will calculate the stability of the bullet. You can select from one of their offerings to have the values auto-filled or manually input them. The G1/G7 boxes deal with the drag function of the bullet. Most commercial bullets are G1, some long range bullets are G7. As noted on the website, you want a stability factor of 1.5 or greater. If you have the details of your bullet design then JBM Ballistics has a calculator: JBM - Calculations - Drag/Twist
When did they stop using musket balls?
It was found that rifles could be made more accurate by elongating the bullet, but the musketball was used for many decades passed the introduction of heavy-for-caliber pointed (a.k.a. spitzer) bullets. Many countries and their militaries carried out test firings from the ,mid eighteenth century ,on large ordnance to determine the drag characteristics of each individual artillery projectile. These findings were later (a century later) formalized for other projectiles — notably the Minie’ of Civil War fame. The concept of stabilizing the flight of a projectile by spinning it was known in the days of bows and arrows when the feathers were offset to produce rotation, but early firearms using spherical balls had difficulty with rifling because of the small area of contact a ball has with the rifling. For a spherical ball or a elongated one, the proper spin helps stabilize its flight. It is only necessary that the poles of the rotation remain pointed along the trajectory of the projectile. Maintaining the proper rotation is more easily assured with an elongated projectile. Once the bullet starts to “yaw” with respect to the poles of rotation, any hope of accuracy is lost — the bullet will begin to veer off in random directions like a poorly struck golf ball or a properly thrown “curve ball” pitch in baseball. Around 1850 ,in England, Francis Bashforth tested rifled artillery guns of 3 inch (76 mm) through 9 inch (229 mm) caliber; and smooth-bore guns of similar caliber for firing spherical shot and howitzers propelling elongated projectiles having an ogive-head of 1½ caliber radius. ,An ogive is the roundly tapered end of a two-dimensional or three-dimensional object. Bashforth then proposed a practical, if somewhat over confidant idea that greatly simplified things, and it is used in the present day. He proposed a model projectile, or “standard” bullet, on which comprehensive measurements of drag deceleration versus velocity could be made. Of course, he had no reason other than optimism that this idea would operate throughout the dimensions of scale from handgun bullets to artillery projectiles. The ballistic tables he introduced were made up for a “carefully created” projectile of "an [artificial] weight and with a specific shape and specific dimensions in a ratio of calibers." For other bullets this “standard” could be scaled by some means, so that exhaustive drag measurements could be avoided for those bullets. He called this the Ballistic Coefficient. The “standard” bullet used in this scheme has an artificially assigned ballistic coefficient of 1.0. A bullet that retains its velocity only half as well as the standard model has a ballistic coefficient of .5. Bullets for small arms have BCs in the range 0.12 to slightly over 1.00. Those bullets with the higher BCs are the most aerodynamic, and those with low BCs are the least. The higher the Ballistic Coefficient of a nonstandard bullet, the lower the drag is compared to the standard bullet. This scale factor turned out to be the form factor of the nonstandard bullet divided by the sectional density. Sectional density (SD) is the numerical result of a calculation that compares a bullet's weight to its diameter. Probably the best way to compare different calibers is by sectional density, not bullet weight. Comparing calibers by bullet weight can be deceiving because a very thin bullet can be made very long. Heavy-for-caliber pointed (spitzer) bullets with a boat tail design have BCs at the higher end of the normal range, whereas lighter bullets with square tails and blunt noses have lower BCs. The higher the BC of a bullet, the better will be its ballistic performance. Ballistic coefficients of most sporting and target bullets have values less than 1.0, and generally BC values increase as caliber increases. Many large caliber bullets have a BC higher than 1.0. In the 1880s, an Italian Army team led by Col. F. Siacci formulated an analytical approach and tied form solutions to the equations of motion of bullet flight for level-fire trajectories. The most commonly used drag model is the G1 model, which is based on a flat-based blunt pointed bullet. This approach meant that trajectory calculations for shoulder arms could be performed algebraically, rather than by the more tedious methods of calculus. Col. James M. Ingalls of the U.S. Army Artillery adopted the Siacci approach. His team produced the Ingalls Tables, first published in 1900, which in turn became the standard for small arms ballistics used by the US Army in World War I. Between about 1865 and 1930 many firing tests were conducted almost worldwide to determine the drag characteristics of standard bullets adopted by different countries. Of particular note were tests made by Krupp in Germany in 1881 and by the Gavre Commission in France from 1873 to 1898, although many other tests were made. The Gavre Commission work was very comprehensive, including not only extensive firing tests going up to a velocity of 6000 fps but also a comprehensive survey of data available from tests in other countries. The Commission attempted to correlate all these data, and it published a composite drag characteristic for a certain standard bullet configuration. The test data turned out to be the basis for ballistics tables for small arms, especially sporting and target ammunition, right up to the present time. For a further discussion see:
What would the bullet drop be at 1000m for an average (World War era) 8x57 Mauser round?
Short answer: Approximately 549 inches or 1,395 centimeters assuming a 100 yard zero. Long answer: The Mauser K98k typically fired the “7,9mmsS” round, a 197gr bullet with a G1 Ballistic Coefficient of .547 at approximately 2500fps., According to ShootersCalculator.com, the ballistic solution for that combination at sea level on a day of average temperature is as follows: Drag Function: G1 Ballistic Coefficient: 0.547 Bullet Weight: 197 gr Initial Velocity: 2500 fps Sight Height : 1.5 in Shooting Angle: 0° Wind Speed: 10 mph Wind Angle: 90° Zero Range: 100 yd Corrected For Atmosphere, Adjusted BC: 0.547 Altitude: 0 ft Barometric Pressure: 29.92 Hg Temperature: 60° F Relative Humidity: 50% Speed of Sound: 1117 fps
What cartridge has the highest known ballistic coefficient?
What cartridge has the highest known ballistic coefficient? Well the direct answer to the question, (as no further info was requested, and assuming factory loads) is the Hornady A-Max 750gr .50 Browning Machine Gun cartridge developed for target/match shooters and has a ballistic coefficient of 1.05 on G1 drag model. While BC has become a buzzword of sorts in the rifle cartridge world, it really is just a measurement of how well the bullet performs in flight against atmospheric variables. The bullet itself is only part of that equation. Bullet designs have either been modified or created to reduce drag, minimize wind drift, ensure stabilization, maintain velocity and retain energy downrange. These factors equal consistency, and consistency equals accuracy. Bullet concentricity, shape, and material all play a vital role in how well it cuts through the atmosphere around it. Wind, pressure, humidity etc all affect a bullet in flight, so the objective is to slice through these variables as cleanly as possible. Bullet weight and velocity determine its ability to mitigate outside forces as well. The other components that are rarely mentioned but are known by ballisticians to be equally important are the ,case ,and ,powder. ,Ballistic ,efficiency, is directly related to propellant. It’s important to note that the same amount of Reloader 25 and IMR 7828 do not burn at the same rate, thus impacting the bullets velocity, trajectory, stability etc. from muzzle to target. Further more, just adding more velocity doesn’t mean better efficiency. If you load a hot long cased magnum cartridge with fast burning propellant, you will get nowhere near the BC you could by using a slower burning powder. This is where case design comes in. Rifle case designs alter the burn rate, compression, and efficacy of acceleration through the rifles grooves and trajectory to target. The overall case length, shoulder slope, angle of the neck all play a vital role in how the powder column burns. The ,case, actually dictates a projectiles efficiency. A longer case can be packed with more propellant, which increaces velocity, but almost inevitably reduces efficiency. Why? Well take the 7mm Mag vs 7mm WSM. A long 7mm magnum case leaves a lot of unburned powder whereas a WSMs shorter wider case packs the powder closer to the primer and burns quickly and evenly before the bullet leaves the case. This allows the shorter cased rounds to get the same velocities with less recoil, muzzle blast and with less powder. The velocity per grain of powder is almost identical, and with the right powder and bullet, a higher BC is achieved. We can see more examples of case + powder + bullet design = higher efficiency and BC in modern cartridges designed around modern powders. The 6.5 creedmore for example, is a shorter, wider case with sharp shoulders, compacted propellant and a long, spire shaped bullet. The BC is very high and many long range shooters have switched over to it because of its ability to fly farther, faster, and straighter than say it’s .308 counterpart. Typical reduction in wind drift side by side is +\- 30%! If you have ever shot long range you know the difference between 9 mils of wind at 3:00 vs 5–6 Mils… If not, that’s about 9–12 feet less drift at 1000 yds. Furthermore, the .308 drops about 32 feet at 1000yds vs 23–25 feet from a 6.5 creedmore. So that’s 7–9 feet less drop with 9–12 feet less wind drift. In long range shooting that’s like comparing a laser beam to an artillery shell… So ,Ballistic Coefficient ,is directly related to ,Ballistic Efficiency. ,More efficient cartridges are typically more accurate because there is less bullet tip-off when the projectile exits the muzzle, and they are more easily fired because there is less recoil and muzzle blast. In addition, high velocity bullets reduce barrel life, and the length and rate of twist in a barrel greatly impacts velocity. A hot loaded long cased inefficient cartridge firing a 200gr projectile through a 30”barrel with a 1:9 twist is going to melt the bullet and destroy the grooves within an afternoon. Even if the ,bullet, ,has a high ,Ballistic Coefficient, ,it would never be realized because of the ballistic inefficiency from the cartridge as a whole.
Why is 7.62mm the choice ammo for top tier sniper rifles (-50bmg)? Why not 5.56mm?
This is a physics problem. If you have both a heavy and a light weight travelling at the same speed, which has more force? The heavy object does. What happens if he light object has more drag than the heavy object? The heavy object stays in motion longer than the light object. The same thing happens with bullets. The 5.56 M855 is a 62 grain bullet and has a muzzle velocity from an M-4 carbine at about 2,900 fps. It’s G-1 ballistic coefficient is 0.274 The 7.62 M80A1has a 130 grain bullet that exits the 24″ barrel of a sniper rifle at about 2,600 fps and has a G1 ballistic coefficient of about 0.365. This means that despite a slightly lower muzzle velocity, the 7.62 will travel farther and carry more energy. The 5.56 will go subsonic at about 700 yards and will only have about 174 ft/lbs of energy at that point. The 7.62 will go subsonic at about 850 yards and will still have 344 ft lbs of energy left. However, snipers generally use handloaded ammunition with a 175 grain bullet with a BC of 0.505 that exits the rifle at just a shade under 2,600 fps. That means it won’t go subsonic until about 1,200 yards and will still have 452 ft/lbs of energy.
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