[guns'ahoy]

from a bench rest, with the barrel pointed level to the ground .. what kind of bullet drop can be expected from a .45 or 9mm after 25 yards? (i know drop will be different for each, im kindly asking for both) ...

[Mal H]

You tell me what velocity for each bullet and I'll plug it into the formulas. We'll assume standard weight FMJ for each, because I don't think the ballistic coefficient will make a lot of difference in that short of a distance. In fact, the vel is the most important factor whether you're talking 9mm or .700 NE.

For some nominal data:
.45 ACP 230 gr FMJ at 850 FPS drop at 25 yds = 1.52 inches
9mm 125 gr FMJ at 1150 FPS drop at 25 yds = 0.85"

[guns'ahoy]

You tell me what velocity for each bullet and I'll plug it into the formulas. We'll assume standard weight FMJ for each, because I don't think the ballistic coefficient will make a lot of difference in that short of a distance. In fact, the vel is the most important factor whether you're talking 9mm or .700 NE.

For some nominal data:
.45 ACP 230 gr FMJ at 850 FPS drop at 25 yds = 1.52 inches
9mm 125 gr FMJ at 1150 FPS drop at 25 yds = 0.85"

thanks... that's basically what i was looking for.. could you plug those same figures in for 100 yards ? (and include a .40sw as well ?)

i

[Mal H]

Sure:

.40 S&W 180 gr FMJ at 1000 FPS drop at 25 yds = 1.1"

.45 ACP 230 gr FMJ at 850 FPS drop at 100 yds = 25.2"
.40 S&W 180 gr FMJ at 1000 FPS drop at 100 yds = 19.0"
9mm 125 gr FMJ at 1150 FPS drop at 100 yds = 14.8"

[10-Ring]

What's the forumla so I can figure those drops myself?

[New_comer]

I believe that the formula for free-falling bodies still apply:

h= -1/2 x g t^2

where
g=gravity constant,
t=time in flight,
h=drop.

Adjust for either metric or english system.

[larryw]

There are several online ballistic calculators, such as http://www.handloads.com/calc/ that will do the math for you. Trick is getting the ballistic coefficient of the bullet, but you can usually get a number close enough for most purposes from the online reloading data published by bullet companies like Speer, Hornady, etc.

[Intune]

I have a question. Is a big open hollowpoint just terrible at cutting through the air compared to a pointed fmj? Let's say for an equal weight & load .45? I would think the diff would be huge but I've been wrong before. Once.

:rolleyes:

[larryw]

Don't have data handy for 45, but comparing the BC of a Speer .429" dia. 240gr JHP to a TMJ SIL (kinda pointy for a handun bullet), the pointy one has a BC that's about 25% better (.165 vs .206 - higher is better). When you run the numbers on these two bullets at magnum loading, the pointy one retains a bit more velocity (and energy as that's a function of the square of velocity) at all distances. However, this does not translate to significant improvements in how flat it shoots until we get to 250+ yards (where drop is four feet vs. five feet).

[Mal H]

larry's right, a ballistic calculator is the best thing to use since it takes into account the BC, elevation (both barrel angle and elev. above sea level), etc. I used the Sierra calculator "Infinity" for the above calcs.

Intune, intuitively I don't think the hollow point is going to cause as rapid a decrease in velocity as you might think. As I implied, that statement is based on intuition and some chats with Harold Edgerton a while back. If you look at some spark gap or high speed flash photos of bullets, there is a shock wave that precedes the bullet and acts as the bow, so to speak, such that the frontal shape of the bullet makes less difference than it would seem.

Hold on and I'll see what the calculator has to say for a FMJ vs. HP for a same weight and vel bullet ...

working ...

working ...

I'm back. That didn't take long did it? ;)

I selected two Speer 230 gr bullets, a Gold Dot HP and a TMJ. Range is 100 yards to emphasize the drop in velocity.

230 GDHP: muzzle vel = 850 FPS, vel at 100 yds = 772.5 FPS
230 TMJ: muzzle vel = 850, vel at 100 yds = 781.0 FPS

That's only about a 9 FPS difference over 100 yards which is less than you'll probably get as the velocity spread for a group of bullets of the same type.

[bountyhunter]

You need to calculate the time to the target:

X = v x t so t = X/v

Where "X" is the distance to the target, "v" is the average bullet velocity. make sure both are in the same units, ie "feet" and "feet per second".

Then, to see how far it drops:

d = 1/2 G x t (squared)

where "G" is the gravitational acceleration constant of 32.2 feet/sec/sec

and "T" is the time calculated for the bullet to travel to the target.

Note the distance drop "d" will be in feet, not inches. Multiply X12 to get inches.

[New_comer]

If we simplify this with, say, very little loss of speed over 100 yds:

with
d = drop in inches
Y = distance of target in yards
MV = muzzle velocity

d = 1/2 x 12 in/ft x 32.2 ft/s^2 x (Y x 3 ft/yd / MV ) ^2

d = 193.2 x (3Y / MV)^2

d = 1739 x (Y / MV)^2

Checked it, and holds quite true for Mal H's numbers. :)

But if greater precision at whatever distance is desired, within limits of course, and the decrease in bullet speed over a set distance is known, we may have to use this statement instead:

d = 6955 x (Y / (MV x (2 - Y(1-Q)/d')))^2

where:

Q = ratio of retained speed at a test distance over the muzzle velocity; so if the bullet's velocity after 100 yds (test distance) is 900 fps with a muzzle (initial) velocity of 1000 fps, the Q would be 0.9 (900 divided by 1000).

d' = test distance in yards


The drop would be a teeny bit more due to correction caused by reduction of effective bullet velocity. And, should the Y-Q-d' variable reach 2.0, error would result.


I'm trying it with different speed-distance combinations, and it yields results quite consistent with the earlier samples.


And it's really fun!! :D:D:D