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In a word yes, especially when coupled with velocity. As for wind reading, it's much easier to do after the last one lands.
Check out the serious extreme distance shooters and see what they are using.

Them aint my rules, but they are the rules if your serious about shooting those kind of distances.

Do bullets with same BC & velocity but different weight/caliber drift the same?

Bullets having the same BC and MV will have the same trajectories; drop, drift, retained velocity, tof, etc. The only difference will be energy.

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.

I could be wrong, but didn't you say "the ability to easily spot shots through the scope" when discussing shooting your .243AI/105AM at a mile?

Fhuqking Texans crack me up...especially when doing their BEST.................(grin)

Fhuqking Texans crack me up...especially when doing their BEST.................(grin)

I also expect smaller diameter bullets to be not quite as well balanced as larger diameter bullets. I found this to be a general tendency when fooling with a Juenke Concentricity Compatator tool years ago, and it makes sense, as a .0001 variation in diameter (or jacket thickness) is comparatively more in a smaller caliber bullet than a larger caliber.

I never have tried to shoot over 500 yards with a .223, and I don't know what a 224 looks like, please remember I'm doing this for the fun and to learn something, it's not life or death to me. Rio7

One of the reasons Ol' Elmer gave for favoring larger bores way back when. Some things are slow to change, eh?

Trybone,

Myths and Wives Tales die lingering deaths,simply because Joe Average ain't very fhuqking bright...as you obliviously attest. Hint. Congratulations?!?

Bless your heart.

Hint.

Laughing!...........