Do bullets with same BC & velocity but different weight/caliber drift the same??
- Thread starter calgarycanada
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mikecr said:
Or just re-read Bryan's first post in this thread. Here:
"Bullets having the same BC and MV will have the same trajectories; drop, drift, retained velocity, tof, etc. The only difference will be energy."
The reason that this is true is because bullet sectional density (not just mass) is part of the BC calculation.
And ending a post in this forum with "IMO" isn't useful, because opinions don't determine trajectories -- physics does.
i will defer to others about the BC-velocity-same performance of each bullet
The primary reason i have for shooting a 30cal bullet vs. the 223 are two fold
1) 30 cal bullets are 27% larger in dia then a 223. and (to me) is an important consideration when scoring on an F class targets if you cut the line with a .30 you may not touch the line with a .223. The rules state that each caliber should be treated the same (in Canada) but there is too much guess work in estimating what .30 cal dia would be...so each caliber is scored, as shot.
2) this has nothing to due with bullet performance and everything to do with scoring. I have seen more then a few times when the target comes up with a MISS because the scorer can not find the small .223 hole amongst 2 dozens patches or because the bullet is outside of the 8 ring and not visible to the scorer.
I shoot with a group of 223 shooters in Ottawa and they are as good at calling the wind as the rest of us but have greater scoring difficulties then us big .30 cal shooter. hopefully Manitou will post on here about his scoring frustrations and benefits of 223 (he is an accomplished 223 shooter.)
Trevor
The primary reason i have for shooting a 30cal bullet vs. the 223 are two fold
1) 30 cal bullets are 27% larger in dia then a 223. and (to me) is an important consideration when scoring on an F class targets if you cut the line with a .30 you may not touch the line with a .223. The rules state that each caliber should be treated the same (in Canada) but there is too much guess work in estimating what .30 cal dia would be...so each caliber is scored, as shot.
2) this has nothing to due with bullet performance and everything to do with scoring. I have seen more then a few times when the target comes up with a MISS because the scorer can not find the small .223 hole amongst 2 dozens patches or because the bullet is outside of the 8 ring and not visible to the scorer.
I shoot with a group of 223 shooters in Ottawa and they are as good at calling the wind as the rest of us but have greater scoring difficulties then us big .30 cal shooter. hopefully Manitou will post on here about his scoring frustrations and benefits of 223 (he is an accomplished 223 shooter.)
Trevor
Newton's 2nd law of motion states the heavier has a higher inertia so the drag will have a lessor effect lol
Trevor60 - yes, unfortunately all too true, especially if the target centre is heavily patched. I had it happen once in the last shot of a regional competition shot over three distances. I spoke to the marker afterwards (a friend and very experienced TR shooter) who told me he saw the bullet hit the backstop sand 'right where it should be', but nobody could find a hole on the paper no matter how hard they searched. (The match organisers had been ridiculously 'tight' on provision of F-Class target centres expecting a single centre on each of the three frames to last all day, so they were a mass of patches by this stage.)
I think when choosing a caliber, you have to be realistic with how far you'll be shooting and YOUR ability. A friend who was shooting a 223 at 600 yards only, thought a 308 would gather more points. He was very wrong. He spent thousands building a great rifle but after developing loads with the 185's and getting great results on paper at 200 yds off a bench, he could not shoot the 308 better (and most times worse) than the 223 at 600.
The 308 is tougher to manage when dealing with recoil, position, and inexperience.
Sometimes bigger isn't better,
RB
The 308 is tougher to manage when dealing with recoil, position, and inexperience.
Sometimes bigger isn't better,
RB
Keith Glasscock
Gold $$ Contributor
I'll second that notion..
Went through the BC equals holes on target for several years with the 223/90.
Came at it in every possible way... I had to as my shoulder is toast so can't deal with the 308 recoil at that time
Despite my best efforts, and I got the 223 shooting very very well, you simply cannot overcome INERTIA.
In steady state air, the 22cal 90gr VLD is workable and can drift LESS then the 185. Or close enough, the error in no wind zero is all that separates them. And I mean in wind speeds needing double digits scope adjustment.
In gusty or bumpy air, forget about the 90gr at 1000yds (works very well at mid range though).
A localised gust pushes the lighter bullet far more then the heavier 30cal. Don't care what the math says... or doesn't say. I just watch my target and those beside me.
When we all screw up, I drifted more in both elevation AND windage.
Small bullet... big wind... not a good partnership.
Now, I have figured out how to deal with the 308 recoil... life is much better. But I have no problem dusting off the 223 for mid range work... beat a bunch of 308's over the years so it is very competitive.
Jerry
Came at it in every possible way... I had to as my shoulder is toast so can't deal with the 308 recoil at that time
Despite my best efforts, and I got the 223 shooting very very well, you simply cannot overcome INERTIA.
In steady state air, the 22cal 90gr VLD is workable and can drift LESS then the 185. Or close enough, the error in no wind zero is all that separates them. And I mean in wind speeds needing double digits scope adjustment.
In gusty or bumpy air, forget about the 90gr at 1000yds (works very well at mid range though).
A localised gust pushes the lighter bullet far more then the heavier 30cal. Don't care what the math says... or doesn't say. I just watch my target and those beside me.
When we all screw up, I drifted more in both elevation AND windage.
Small bullet... big wind... not a good partnership.
Now, I have figured out how to deal with the 308 recoil... life is much better. But I have no problem dusting off the 223 for mid range work... beat a bunch of 308's over the years so it is very competitive.
Jerry
Laurie said:
That's been my experience precisely. The 223REM w/ 90 VLDs performs as well as the 185 Juggernaut.....then you get the random 9-ring or 8-ring elevation shot every so often.....more often when it's windy.
skiutah02
Silver $$ Contributor
Wish I would have seen/known this a few months ago with respect to the 90s. Been tearing my hair out trying to get the one flyer (OK sometimes 2) into the group of 10 for a few months (and several hundred rounds). I will say that the pursuit has upped my reloading game...Drew
This is the part I wanted Brian Litz to explain. 90s in 223 vs 185s in 308 is one example, I just wanted to understand the concept of "INERTIA" and WHY (if it exits) it's not part of BC and so many advanced ballistic programs. 223 case has its limitations so let's leave it out. Let's say 6mm 105s(G1 .547) and 30cal 185 (.549)VLD shot at same velocity, would 30 cal drift less in 'twitchy' winds? All ballistic programs available to general public will tell me they will drift the same(almost), including the ones developed by Bryan himself. BC of these bullets is very accurate so let's not argue over 'real BC might be different'. So is he wrong? Brian if you can please chime in that'll be great.mysticplayer said:
Sorry guys not trying to fuel the argument (I apologize if it sounds that way) I'm just trying to understand science of ballistics and its limitations.
It does exist and is part of bc, you just have to dig a bit...
G7BC = [bullet weight / caliper2] / g7 form factor
and...
Bullet Weight = bullet mass *grav. acceleration
(Don't use g1 bc.. The g7 form factor is constant.... G1 form factor is Varable!!!)
Inertia increases with the bullet's mass. More mass means more inertia. More mass also means more weight, which results in a higher bc, all things being equal. The bc (along with mv) really does incorporate everything you need to find a basic trajectory.
The problem is that the math becomes less useful if you cross the fringes of performance. What good is comparing bc of 2 different bullets if one of them is unstable at the target?
I'd agree with those comments Jerry, especially as my main shooting venue, Diggle, is in a rising, narrowing valley where winds can have a significant elevation effect alongside the lateral component.
However, I had an even 'better' (worse) example last September in the European F-Class Championship meeting at Bisley shooting a 6.5X47L with the 123gn Scenar at ~3,000 fps in F-Open. Due to a squadding mistake I shot with my FTR friends and could compare what happened to my shots against theirs. It was a week of light winds that also produced many dropped shots on verticals, nearly always low, for almost everybody. The fast 7s almost shot through the effects, the 308s were noticeably affected and my 6.5 bullet was worse still despite straight ballistics program range charts saying that any wind effect should be less for me than a 308 FTR bullet flight.
That's what I believed in but when very respectable shooters like Mysticplayer and Laurie believe inertia is something beyond BC, I developed my doubts. Hence this thread. But I picked wrong example initially coz 90 VLD out of 223 may not be stable by 900 mts? That's why I picked next two bullets with closest BCs(I should have used G7)Joe Grad said:
Bryan Litz Ballistics
Site $$ Sponsor
As stated, inertia is accounted for by BC because the bullet weight is accounted for in BC.
But at supersonic speed, aerodynamics play a much larger role than mass.
Consider that, a .30 caliber 185 grain Juggernaut bullet has 1.26 lb of air resistance at Mach 2.5 (2790 fps), and the .224 cal 90 grain VLD has 0.62 lb of air resistance at the same speed. The .224 cal bullet has so much less air resistance for two reasons; primarily because of it's smaller cross section, and further because of it's lower drag shape.
Now consider that you've got 1.26 lb of drag on a bullet that weighs only 185/7000 = 0.0264 lb; the force of aerodynamic drag is 47.73 times greater than the weight of the bullet. This relationship is proportional to the bullets acceleration (velocity decay).
The .224 caliber 90 grain bullet weighs 90/7000 = 0.0129 lb, so aerodynamic drag on this bullet is 0.62/0.0129 = 48.06 times greater than it's weight.
You'll notice that; although both bullets have drastically different amounts of aerodynamic drag, and both have different mass, the ratio of aerodynamic drag and bullet mass is essentially the same, with the .30 cal being slightly better. In other words, their acceleration (velocity decay) will also be very similar.
You may also recall that the BC's of these bullets are also nearly the same, with the .30 cal 185 being slightly better.
The above is just a different way of looking at the math behind BC, which breaks out the aerodynamic drag from the mass. It shows how/why two bullets with the same BC will slow at the same rate; it's because they experience the same ratio of aerodynamic drag compared to their weight.
For those interested in further explanation, you can calculate the aerodynamic drag in pounds using the formula:
F=1/2*rho*V^2*S*CD
Where:
F is the force of drag in lb
rho is the air density (0.002375 sl/ft^3)
V is the bullet velocity in fps (Mach 2.5 is 2790 fps)
S is the bullet's cross sectional area in ft: (.308/2/12)^2*pi = .000517 for .308 cal, and (.224/2/12)^2*pi = .000274 for .224 cal
CD is the drag coefficient at a given speed (.263 for the 185 Juggernaut at Mach 2.5, and .243 for the 90 grain VLD at the same speed).
Solving for the .30 cal 185:
F=1/2*.002375*2790^2*.000517*.263
F=1.257 lb
Replace variables for .224 to calculate it's drag.
That's actually the hardest part about solving a ballistic equation. Knowing the force, you can solve for the (negative) acceleration. Want to know how much the bullet slows down in 0.1 seconds, Newtons second law is: F=ma (Force = mass times acceleration). Said differently, acceleration is Force divided by mass.
a=F/m
We just calculated the force: 1.26 lb for the 185 grain, acceleration is then: 1.26/(185/7000/32.2*) = 1535 fps/s. (*Note you have to divide weight by 32.2 to get mass in slugs.)
So the bullet should slow down about 153.5 fps in .1 seconds. The ballistics program says: 148.5. Why the difference? Because as soon as the bullet started slowing down, the force of aerodynamic drag began diminishing. In a real ballistics solver, these equations are solved every 0.001 seconds in order to accurately capture the (negative) acceleration of the bullet as it flies downrange.
Further complicating matters is the fact that the drag coefficient changes with Mach number. If you're refering to a standard drag model (G1 or G7), the CD's are tabulated against Mach the same in every ballistics solver so it makes it easy. The G7 standard drag model is a better match for our long range bullets.
To close the loop, we can extract BC from Newtons second law:
a=F/m.
We know the aerodynamic Force from above, so:
a=1/2*rho*V^2*S*CD/m
The inverse of sectional density can be pulled out of the above:
a=1/2*rho*V^2*pi*cal^2*CD/(4*144*m)
a=1/2*rho*V^2*pi*CD*cal^2*32.2/(4*144*bw) with the bold terms being 'SD' terms. Note we removed the 32.2 factor and are now using bw (bullet weight) instead of mass.
Remember the equation for BC=bw/(cal^2*i7)
Where i7=CD/CD7 (drag coefficient of your bullet divided by the drag coefficient of the G7 standard)
Using these terms in Newtons second law:
a=1/2*rho*V^2*pi*CD7*i7*cal^2*32.2/(4*144*bw)
And now the bold terms are BC terms. Note that BC is inverted here which makes sense because higher BC means less acceleration (velocity decay).
Now let's see if the numbers work out. Note CD7 is the G7 drag coefficient, in this case we're looking at Mach 2.5 where it's .270, so i7 is .263/.270 = .974. Note this is the i7 specifically at 2790 fps. The average i7 from 3000 to 1500 fps can be found by dividing the bullets sectional density by it's G7 BC: .279/.283=.986. So the form factor is pretty constant.
Getting back to the verification above:
a=1/2*.002375*2790^2*pi*.270*.974*.308^2*32.2/(4*144*185/7000)
a=1532 fps/s.
This is pretty close to the 1535 fps/s we found above. Maybe some round off error somewhere.
So that's the math. Pretty straightforward. And BTW, once you've solved for acceleration, everything else (drop, wind deflection, etc) is a given.
So why does the 90 grain appear to not actually match performance of the .30 cal 185 in actual match conditions? As I stated in an earlier post, I suspect that stability of the 90 grain bullet is challenged more at low supersonic speed due to it's greater length (lower Ix/Iy ratio), in other words it's more 'tipsy'. So at long range it flies with more yaw and induces more drag. So it's not always flying with it's full potential BC.
Why wouldn't everyone agree on this? Well, shooting at higher altitudes would allow the bullet to reach 1000 yards in better shape (higher stability, less induced drag). So if you're a 1000 yard shooter at 5000 ft altitude, you may not see as big of a performance difference between the 90 grainers vs. .30 cal 185's.
Another reason (and this speaks to the fliers) is that you're spinning a proportionally longer bullet much faster, so dispersion will be exacerbated more. The 90 grain bullet is more likely to have substantial jacket run out due to it's long length, and therefore more imbalance. Spin that puppy in a 1:7" vs. the 185 which is likely better balanced and spinning out of a 1:10", and it's easy to explain the fliers.
Shooters like to draw contrasts between 'theoretical/paper' analysis vs. 'real world' results. The gap can be big if your theoretical analysis is basic, but the more you learn, the more that gap closes.
-Bryan
But at supersonic speed, aerodynamics play a much larger role than mass.
Consider that, a .30 caliber 185 grain Juggernaut bullet has 1.26 lb of air resistance at Mach 2.5 (2790 fps), and the .224 cal 90 grain VLD has 0.62 lb of air resistance at the same speed. The .224 cal bullet has so much less air resistance for two reasons; primarily because of it's smaller cross section, and further because of it's lower drag shape.
Now consider that you've got 1.26 lb of drag on a bullet that weighs only 185/7000 = 0.0264 lb; the force of aerodynamic drag is 47.73 times greater than the weight of the bullet. This relationship is proportional to the bullets acceleration (velocity decay).
The .224 caliber 90 grain bullet weighs 90/7000 = 0.0129 lb, so aerodynamic drag on this bullet is 0.62/0.0129 = 48.06 times greater than it's weight.
You'll notice that; although both bullets have drastically different amounts of aerodynamic drag, and both have different mass, the ratio of aerodynamic drag and bullet mass is essentially the same, with the .30 cal being slightly better. In other words, their acceleration (velocity decay) will also be very similar.
You may also recall that the BC's of these bullets are also nearly the same, with the .30 cal 185 being slightly better.
The above is just a different way of looking at the math behind BC, which breaks out the aerodynamic drag from the mass. It shows how/why two bullets with the same BC will slow at the same rate; it's because they experience the same ratio of aerodynamic drag compared to their weight.
For those interested in further explanation, you can calculate the aerodynamic drag in pounds using the formula:
F=1/2*rho*V^2*S*CD
Where:
F is the force of drag in lb
rho is the air density (0.002375 sl/ft^3)
V is the bullet velocity in fps (Mach 2.5 is 2790 fps)
S is the bullet's cross sectional area in ft: (.308/2/12)^2*pi = .000517 for .308 cal, and (.224/2/12)^2*pi = .000274 for .224 cal
CD is the drag coefficient at a given speed (.263 for the 185 Juggernaut at Mach 2.5, and .243 for the 90 grain VLD at the same speed).
Solving for the .30 cal 185:
F=1/2*.002375*2790^2*.000517*.263
F=1.257 lb
Replace variables for .224 to calculate it's drag.
That's actually the hardest part about solving a ballistic equation. Knowing the force, you can solve for the (negative) acceleration. Want to know how much the bullet slows down in 0.1 seconds, Newtons second law is: F=ma (Force = mass times acceleration). Said differently, acceleration is Force divided by mass.
a=F/m
We just calculated the force: 1.26 lb for the 185 grain, acceleration is then: 1.26/(185/7000/32.2*) = 1535 fps/s. (*Note you have to divide weight by 32.2 to get mass in slugs.)
So the bullet should slow down about 153.5 fps in .1 seconds. The ballistics program says: 148.5. Why the difference? Because as soon as the bullet started slowing down, the force of aerodynamic drag began diminishing. In a real ballistics solver, these equations are solved every 0.001 seconds in order to accurately capture the (negative) acceleration of the bullet as it flies downrange.
Further complicating matters is the fact that the drag coefficient changes with Mach number. If you're refering to a standard drag model (G1 or G7), the CD's are tabulated against Mach the same in every ballistics solver so it makes it easy. The G7 standard drag model is a better match for our long range bullets.
To close the loop, we can extract BC from Newtons second law:
a=F/m.
We know the aerodynamic Force from above, so:
a=1/2*rho*V^2*S*CD/m
The inverse of sectional density can be pulled out of the above:
a=1/2*rho*V^2*pi*cal^2*CD/(4*144*m)
a=1/2*rho*V^2*pi*CD*cal^2*32.2/(4*144*bw) with the bold terms being 'SD' terms. Note we removed the 32.2 factor and are now using bw (bullet weight) instead of mass.
Remember the equation for BC=bw/(cal^2*i7)
Where i7=CD/CD7 (drag coefficient of your bullet divided by the drag coefficient of the G7 standard)
Using these terms in Newtons second law:
a=1/2*rho*V^2*pi*CD7*i7*cal^2*32.2/(4*144*bw)
And now the bold terms are BC terms. Note that BC is inverted here which makes sense because higher BC means less acceleration (velocity decay).
Now let's see if the numbers work out. Note CD7 is the G7 drag coefficient, in this case we're looking at Mach 2.5 where it's .270, so i7 is .263/.270 = .974. Note this is the i7 specifically at 2790 fps. The average i7 from 3000 to 1500 fps can be found by dividing the bullets sectional density by it's G7 BC: .279/.283=.986. So the form factor is pretty constant.
Getting back to the verification above:
a=1/2*.002375*2790^2*pi*.270*.974*.308^2*32.2/(4*144*185/7000)
a=1532 fps/s.
This is pretty close to the 1535 fps/s we found above. Maybe some round off error somewhere.
So that's the math. Pretty straightforward. And BTW, once you've solved for acceleration, everything else (drop, wind deflection, etc) is a given.
So why does the 90 grain appear to not actually match performance of the .30 cal 185 in actual match conditions? As I stated in an earlier post, I suspect that stability of the 90 grain bullet is challenged more at low supersonic speed due to it's greater length (lower Ix/Iy ratio), in other words it's more 'tipsy'. So at long range it flies with more yaw and induces more drag. So it's not always flying with it's full potential BC.
Why wouldn't everyone agree on this? Well, shooting at higher altitudes would allow the bullet to reach 1000 yards in better shape (higher stability, less induced drag). So if you're a 1000 yard shooter at 5000 ft altitude, you may not see as big of a performance difference between the 90 grainers vs. .30 cal 185's.
Another reason (and this speaks to the fliers) is that you're spinning a proportionally longer bullet much faster, so dispersion will be exacerbated more. The 90 grain bullet is more likely to have substantial jacket run out due to it's long length, and therefore more imbalance. Spin that puppy in a 1:7" vs. the 185 which is likely better balanced and spinning out of a 1:10", and it's easy to explain the fliers.
Shooters like to draw contrasts between 'theoretical/paper' analysis vs. 'real world' results. The gap can be big if your theoretical analysis is basic, but the more you learn, the more that gap closes.
-Bryan
Thank you very much Bryan. I really appreciate such detailed explanation and putting this matter to rest. Now I understand actual reasons behind the matter and not just because somebody said so.
Thanks again
Thanks again
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