Maximum height of bullet trajectory?

[davidjoe]

I also agree that the reason why this is a common misconception is, as you identified, the 'drop in inches at the target' that most calculators display. For example, JBM shows the bullet drop at the target as -275 inches at 1000 yards for the data you provided. I believe what a lot of people do not realize is that this is not the 'drop from the apex of the bullet trajectory' but rather 'the difference in height at which the bullet would be, from the center of the target, at the target, if gravity did not exist'.

There is a super straight forward explanation for drop. At 1,000 yards, if you do not touch the scope with a 100 yard zero and hold dead center, the bullet will attempt to drop precisely the 32 moa specified for a TR rifle.

Of course, we don’t mount targets high enough off the ground to ever see this drop play out, and they hit dirt well before the target.

When we add 32 moa to a TR rifle’s 100 yard zero, we have moved the crosshairs aim point close to 30 feet. This doesn’t mean the bullet will rise 30 feet, (because it rises and falls versus just “falling” if you don’t dial) and that is simply the amount that a given cartridge/bullet needs to reach 1,000 yards and it’s still less than one degree of difference.

When Staubach era QB’s threw deep with their natural arms, 40 degrees of departure angle, (over the top row seats could be needed) to reach the end zone. The top row-opposite end zone is never in danger of catching a ball aimed at them.

Interestingly, FG kickers must bring the angle down, and simply wail the snot out of football to have any chance at all of 60 plus yards, which means they clearly would be F-Open, not TR shooters.

 

[davidjoe]

Gravity isn’t as strong as bullet trajectory suggests.

I the absence of gravity, the bullet should stop several miles away because of air resistance alone, and float in the air stationary, in line with the bore, waiting for us to go retrieve it and figure out how to swage and reload it.

In the absence of air (vacuum) but with normal gravity, bullets will never (necessarily) slow down. If one knew MV and bore orientation, the exact place of impact could be predicted. Bullet shape, weight, size, and stability would not matter. Forward, backward, sideways. MV would reign.

Gravity “could be” zero impediment to hitting a target in a vacuum, meaning distance can be infinitely compensated by holdover and become almost irrelevant. A target straight up from gravity or almost such, would be an exception but is also an exception to how we shoot. Gravity’s deceleration of a bullet in a typical arc is negligible.

Looked at this way, air, not gravity, is our challenge. To air, BC is our shield and MV is the sword.

 

[DocUSMCRetired]

What you are looking for is Maximum Ordinate. Some of our engines will calculate this for you.

 

[butchlambert]

What you are looking for is Maximum Ordinate. Some of our engines will calculate this for you.View attachment 1482178

Nice to see you on here.

 

[DocUSMCRetired]

Nice to see you on here.

I lurk on here from time to time.

 

[Straightshooter1]

It's been shown that the maximum Time Of Flight for a human leaping off the ground is slightly less than 1 second. What's the maximum TOF of a batted baseball, to the nearest half second?
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Depends on spin rate and density altitude.

 

[DiffEQ]

There are three kinds of people in the world, those who do math, and those who cannot.

 

[Brians356]

Depends on spin rate and density altitude.

Ok, assume Coors Field, 90 degrees, 20% relative humidity, and whatever possible spin rate produces the highest TOF.
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