Definition of flat shooting.

nessy357

nessy357

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Let's continue this here so that nearnorthguy doesn't get hijacked.

What is your definition of "Flat shooting ?"

e.g. My 50 cal shoots flat out to 75 yds with 55 grns of FFFg,
It shoots flat out to 100yds with 65 grns of FFFg.

Cheers nessy.
 
Well as they all shoot in an arc flat shooting may be the wrong term but we all know what you mean my 300ultra will shoot to point of aim to 300yds but the 577 only 100yds .
 
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My definition would be a projectile that has the least amount of arc in its flight.
For example: when shooting @ 100 yards bullseye, with dead on POA, how small is the distance that the projectile shoots high @ say, 50 yards.
A 128 gr round ball will shoot flatter than a 250 grain bullet if pushed by identical powder charges.
I'm guessing, but the ball may be 4" high at 50 yds and the bullet may be 6" high at the same distance to obtain the same POI at 100yds. Hence, the lighter weight ball is "flatter shooting".

Nessy's load of 65gr shoots flatter than the same ball pushed by 55gr., hence the higher POI at 100yds.
 
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Smokepole said:
How much experience do you have in this matter?.
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I own an .50 now, and owned a .54, no .45.

I can read, read a chart, and do math though, having taken university level physics. We also did alot of projectile problems, gravity being so important, and they like to justify us learning vectors.

Smokepole said:
Believe me, you will notice the difference in trajectory between a 128 gr .440 RB and a 177 gr, .490 RB, given the same powder charge.
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I could talk about the fact that a 80 grain (usual factory max) charge of powder does push the 128 .440 ball 22% faster than a 90 grain (usual factory max) charge pushes the 177 .490 grain ball.

So, 22% huh? Not much eh?

Even less. Because although the .490 ball is 38% heavier than the .440 ball, the difference between the two in surface area is only 11.5%. This gives the .490 ball a substantially greater aerodynamic advantage, which means that the 22% velocity advantage of the .440 melts away making the two firearms pretty similar in "flatness" of shooting at reasonable yardage.

I could talk about all that, but I'm not sure you'd have any respect for all that "book learnin". I'm starting to think that this is pointless, that I'm tilting at windmills, and I think I'll make this my last post for awhile, and go back to my "silly" schoolbooks.

What I can't understand though, is your fascination with "THE SAME POWDER CHARGE"

Do you own a single, non-adjustable powder measure for all your guns? Some favorite spoon, you carry everywhere? Is that it? Why do you keep demanding that we compare "the same powder charge" for two different calibers? It really makes no sense to me.

Larger calibers eat more powder, and that seems obvious.

By your logic we should be shooting muzzloaders with .177 caliber BB's loaded over 90 grains of FFF. We'll call the new gun smokepole's white lightning express rifle. It will take six hours to trickle the powder down the bore, and be nine feet long, but DAMNED if the trajectory won't put the 22-250 to shame!:eek:
 
Jeash! I didn't mean to make ya mad, Bobby. For those who missed it and are wondering why he seems choked, the original theread where this started can be found here - http://www.canadiangunnutz.com/forum/showthread.php?t=287176

After a quick look at your figures, I think you have forgotten to calculate time of flight into your figures.
I may not have gone to University, but if the velocity would remain constant for the entire time of flight, the .45 cal round ball would still drop 22% (according to your figures, Bobby) less than the .50 round ball, even with the heavier powder charge you have suggested. This is due to the 22% less time that gravity has to act on it. This is not accurate in a real world scenario though, as Bobby has pointed out the heavier .50 ball will slow down less than the .45 ball will in the same distance. By what percentage I couldn't tell you.

I have suggested the same powder charge for two different calibers to set the variables. If we try to compare two completely different loads the figures mean little. For example, we could use a round ball in a .50 and a 350 gr bullet in the .45. I don't need math to know that the .50 RB will shoot flatter than the .350gr .45 bullet even if you raise the charge 10 - 20 gr behind the heavier bullet.
All in all, I conceede to you Bobby, as you have pointed out that by varrying the load one can make a .50 shoot as flat as a .45.

Now Bobby, since you have experience with a .50 and a .54, wouldn't you say the .50 was a flatter shooting caliber? He he... evil laugh... lol.
 
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Bobby telling every one that you have a university level physics degree just sounds like your blowing your own horn . and as this is antique forum it probably doesn't fly here people here may have a different term for it it's trial and error and mostly don't care why it works just that it does no matter what the math . You may notice I don't go posting I'm a Technologist it just don't matter.Try to keep the fun in it some people just take themselves too serious .
 
Bobby telling every one that you have a university level physics degree just sounds like your blowing your own horn . and as this is antique forum it probably doesn't fly here people here may have a different term for it it's trial and error and mostly don't care why it works just that it does no matter what the math . You may notice I don't go posting I'm a Technologist it just don't matter.Try to keep the fun in it some people just take themselves too serious .
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I thought it was pretty swell of Bobby to take the time to type a page of his knowledge.

As for his degree guess what, (for you trial an error guys) It takes a Pile of work and effort to get a degree. He should be proud of it and it does make his Opinions/explainations a lot more valid then what we usually get. Why would anyone want to waste thier time with trial and error when you could use the knowledge contained in the head of your buddy the physics major????

Belittling and undervaluing the PRACTICAL APPLICATION of academic knowledge is rampant in north america right now it is making us the laughing stock of the world.


Bobby Get a law degree to go with that Physics degree and get ready to make some serious coin.

BTW the question posed is pretty lame but bobbie's post was interesting.
 
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Smokepole said:
the .45 cal round ball would still drop 22% (according to your figures, Bobby) less than the .50 round ball, even with the heavier powder charge you have suggested. This is due to the 22% less time that gravity has to act on it.
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Not exactly, since gravity causes acceleration, so half the time is less than half the drop.

The trajectory is flat under zero gravity, so the term has no real meaning here on the ground.
 
Juster said:
Not exactly, since gravity causes acceleration, so half the time is less than half the drop.
QUOTE]

Please explain. As far as I'm aware, gravity can only provide a pull towards the center of the earth. Wouldn't that mean the acceleration is down, causing the curve of trajectory in the projectile's path? A change in the direction of velocity is an acceleration (centripetal acceleration) in physics, IIRC, as is a slowing of velocity (deceleration, in common terms).

So, is this statement correct? A projectile traveling downrange on a path that starts parallel to the surface of the earth will fall at the same rate as an identical projectile dropped perpendicular to the earth's surface, from the same height as the muzzle, as the former projectile.
In other words, drop a bullet from barrel height to the ground at the same instant as one is fired down range. Both bullets are pulled by gravity at the same rate and hit the surface of the earth at the same time.
True or false?
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Juster said:
If a ball drops 10" inches in 1 second, how far will it drop in 2 seconds?
Not 20". It will be more than that because the speed of the falling object is increasing.
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Now that is something I never considered. That is relevant in the last part of a bullet's flight, and explains the greater curve in trajectory further downrange.
My knowledge of physics is limited, so that brings to mind another question:
Does a heavier object accelerate towards earth at a greater rate than a lighter object?
I'm thinking it must and that would explain (partialy) why a heavier bullet has a greater curve in the arc of its trajectory.

Sooo, in Nessy's original posting, the 65gr load shoots higher at 100 yards than the 55gr load because of the less time in flight? Is that the only reason, or is there more to it that I haven't considered?
(I'm thinking the guys over at prescission rifles have hashed this all out already...)
 
That is relevant in the last part of a bullet's flight, and explains the greater curve in trajectory further downrange.
Yes indeed.
Does a heavier object accelerate towards earth at a greater rate than a lighter object?
No, Apollo-15 dropped a hammer and a feather on the moon. They hit the ground at the same rate due to no air resistance. Galileo dropped a ten-pound weight and a one-pound weight off the Leaning Tower of Pisa, they hit the ground at the same time.
shoots higher at 100 yards than the 55gr load because of the less time in flight?
Less time in flight is the only reason for less drop.
 
If the heavier bullet doesn't accelerate more quickly towards the ground due to gravity than a lighter one and time of flight is the only factor in how far a bullet drops form the bore-line... aww, crap! brain meltdown.

Why was there a feather on Appolo 11?
 
Smokepole said:
If the heavier bullet doesn't accelerate more quickly towards the ground due to gravity than a lighter one and time of flight is the only factor in how far a bullet drops form the bore-line... aww, crap! brain meltdown.

Why was there a feather on Appolo 11?
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You guys are all missing the biggest factor that affects the flight of a ballistic projectile and that is its ballistic coefficient. The lower the BC the quicker friction from passing through the air slows the projectile and the greater the arc. gravity is a constant through out the bullet flight and it is only because we compensate for it that the bullet actually crosses the line of sight twice creating an arc that's apex is higher than the end of the barrel. Were we not to correct the bullet would never cross the line of sight and in fact falls from the moment it leaves the barrel.
 
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