Maximum height of bullet trajectory?

[marine52]

You know what, you are both correct! The ambiguity of the question is what allows this to be the case.

These are two more precise questions that will illustrate:

- After the bullets leaves the muzzle, does it ever rise above the plane at which the muzzle is pointing?
The answer is No. This is the question Butch answered and he is correct.

- After the bullet leaves the muzzle, does it ever rise further above the ground than the height at which it leaves the muzzle?
The answer is Yes. This is the question marine52 answered and he is correct.

The key is to make the question explicit and specify 'rise relative to what?'.

Relative to the plane at which is fired - No - because of gravity
Relative to the ground - Yes - because of launch angle

Thanks. I believe this clarification is correct. however, there are many shooters who believe that the bullet is “falling” the instant it departs the muzzle. This is only true in the trivial classroom exercise of a barrel pointed horizontal to the earth’s surface. If the bullet started it’s fall immediately after muzzle departure then the gravitational acceleration factor g (32.17 feet per second per second) would mean that it would be falling faster and faster every second by the g acceleration. However, at departure the bullet has a positive vertical velocity. It is true that gravitational acceleration is acting on the bullet to slow its upward vector and this downward acceleration acts to decrease the bullet’s positive verticle velocity until it reaches the max ord, where, for a brief moment, the verticle velocity is zero. At that point the bullet starts it’s decent at the g acceleration rate.

It is noteworthy to remember that the max ord also corresponds to the midpoint of the time of flight (TOF). This is a simple HS class physics problem of an object thrown up and in free fall. the total TOF of the object thrown straight up will equal twice the time it takes the object to reach its high point, discounting any air resistance, which is reasonable with a projectile or object thrown 10 to 30 feet up

 

[Rat Rifle]

For my 308 win FTR rifle with a 200 grain Berger , the max ord for a 1000 yard target is about 9.5 feet. It all depends on your MV, ballistic coef, and air density (density altitude).

OK. So, is it safe say, given Your 308 example, shot on a level range, that the maximum arc height is “approximately” 30% of the total bullet drop ?

Thanks in advance

 

[marine52]

OK. So, is it safe say, given Your 308 example, shot on a level range, that the maximum arc height is “approximately” 30% of the total bullet drop ?

Thanks in advance

Hard to say. I haven’t done the calculation. To find the true “bullet drop”, defined as line of departure minus trajectory at a point on the trajectory curve, you must calculate the corresponding point on the line of departure. This can easily be done using trig but I don’t know of a ballistic program that does it.

 

[marine52]

OK. So, is it safe say, given Your 308 example, shot on a level range, that the maximum arc height is “approximately” 30% of the total bullet drop ?

Thanks in advance

BTW, Most shooters use drop at distance to compare rifles/cartridges. They have their rifle/ballistic app zeroed at 100 and do the calculation for gravity drop at 800 or 1000 yards for example to estimate and compare “bullet drop.” Is this meaningful? Maybe, depends on your perspective.
To me, the entire use of the “line of departure“ (fake line in the sky) for ballistic talk is confusing. Bullet path or trajectory is what counts. But that’s just my totally biased perspective.

 

[Webster]

Thanks. I believe this clarification is correct. however, there are many shooters who believe that the bullet is “falling” the instant it departs the muzzle. This is only true in the trivial classroom exercise of a barrel pointed horizontal to the earth’s surface. If the bullet started it’s fall immediately after muzzle departure then the gravitational acceleration factor g (32.17 feet per second per second) would mean that it would be falling faster and faster every second by the g acceleration. However, at departure the bullet has a positive vertical velocity. It is true that gravitational acceleration is acting on the bullet to slow its upward vector and this downward acceleration acts to decrease the bullet’s positive verticle velocity until it reaches the max ord, where, for a brief moment the verticle velocity is zero. At that point the bullet starts it’s decent at the g acceleration rate.

It is noteworthy to remember that the max ord also corresponds to the midpoint of the time of flight (TOF). This is a simple HS class physics problem of an object thrown up and in free fall. the total TOF of the object thrown straight up will equal twice the time it takes the object to reach its high point, discounting any air resistance, which is reasonable with a projectile or object thrown 10 to 30 feet up

There is no mechanism to make the bullet rise when it leaves the bore. It's all BS.

 

[Highpower-FClass]

BTW, Most shooters use drop at distance to compare rifles/cartridges. They have their rifles zeroed at 100 and do the calculation for gravity drop at 800 or 1000 yards for example to estimate and compare “bullet drop.” Is this meaningful? Maybe, depends on your perspective.
To me, the entire use of the “line of departure“ (fake line in the sky) for ballistic talk is confusing. Bullet path or trajectory is what counts. But that’s just my totally biased perspective.

Bullet drop is actually a very important measure, but not for determining / knowing the apex of the bullet trajectory.

The problem is when bullet drop is expressed in inches. Knowing the bullet drop in inches is of little use and a source of confusion.

However, knowing the bullet drop in either MOA or MIL is everything as this tells us the comeups for our scopes at a specific distance from the target.

Take this output from JBM ballistics which include both the bullet drop in inches, and the bullet drop in MOA in columns 2 and 3 respectively (this is for the 180 hybrids scenario in the original post):


From this we see that the bullet drop in inches at 1000 yards is 275.4 inches (and yes, the table above is for a 100 yard zero, so plus / minus an inch perhaps).

This is pretty much useless information. It has nothing to do with the trajectory of the bullet, it is simply 'how many inches above the center of the target should the muzzle of the rifle be pointing (when intersecting the vertical plane at 1000 yards) in order for the bullet to arrive at the vertical center of the target?'. We don't dial our scopes in inches (luckily, as that would be a nightmare) so the actual practical use of knowing the number 275.4 inches is negligible.

However, in the next column is the same bullet drop expressed expressed in MOA. This is really useful information as it tells us how much we need to dial up on a MOA scope from a 100 yard zero (in this case) in order to shoot at the vertical center of the target from 1000 yards. In this case that is 26.3 MOA from the 100 yard zero.


Bullet drop in inches - not a very useful piece of information

Bullet drop in MOA or MIL - really good information

 

[marine52]

Bullet drop is actually a very important measure, but not for determining / knowing the apex of the bullet trajectory.

The problem is when bullet drop is expressed in inches. Knowing the bullet drop in inches is of little use and a source of confusion.

However, knowing the bullet drop in either MOA or MIL is everything as this tells us the comeups for our scopes at a specific distance from the target.

Take this output from JBM ballistics which include both the bullet drop in inches, and the bullet drop in MOA in columns 2 and 3 respectively (this is for the 180 hybrids scenario in the original post):

View attachment 1480246
From this we see that the bullet drop in inches at 1000 yards is 275.4 inches (and yes, the table above is for a 100 yard zero, so plus / minus an inch perhaps).

This is pretty much useless information. It has nothing to do with the trajectory of the bullet, it is simply 'how many inches above the center of the target should the muzzle of the rifle be pointing (when intersecting the vertical plane at 1000 yards) in order for the bullet to arrive at the vertical center of the target?'. We don't dial our scopes in inches (luckily, as that would be a nightmare) so the actual practical use of knowing the number 275.4 inches is negligible.

However, in the next column is the same bullet drop expressed expressed in MOA. This is really useful information as it tells us how much we need to dial up on a MOA scope from a 100 yard zero (in this case) in order to shoot at the vertical center of the target from 1000 yards. In this case that is 26.3 MOA from the 100 yard zero.


Bullet drop in inches - not a very useful piece of information

Bullet drop in MOA or MIL - really good information

Yes I agree. The angular value is what counts. It is based on the trajectory of a 100 yard zero and gives you MOA come-ups for shooting at distance. As I said, trajectory is what counts

 

[xr650rRider]

 

[marine52]

That’s some serious elevation on that barrel. Jerry Miculek is a fantastic shot.

 

[Webster]

(This has been discussed somewhat before, but I have not seen a definitive answer. )

Numerous times I have heard conversations regarding how high our bullets fly on their way to the targets. Some shooters opined that many of the flags that are 10 to 12 feet high were useless since the bullets reach 20 to 30 feet altitude. But looking at our rifles when we shoot, even at 1000 yards, I could not imagine any bullet going 30 feet high since the rifles are essentially level. I believe the error is that they assume that the peak of the trajectory is the same as the bullet drop at the target, and I believe this is incorrect.

Using the Berger Ballistic Calculator, I used 0 as the starting range, and 1000 as the ending range, and plotted the results for 180 gr Hybrid Target bullets at 2825 fps muzzle speed. These show the peak height is at around 500 yards down range, and is closer to 89 inches, or 7-1/2 feet:

View attachment 1479967

What do you think?

Thanks!
Alex

The chart is misleading. The ht. as referenced to what. It's simple, draw a picture. Just remember the line of sight crosses the bullet trajectory. 26" high at 100 yards! Not sure what the chart represents. It isn't a drop chart with the gun zero'ed at the muzzle.

 

[Webster]

(This has been discussed somewhat before, but I have not seen a definitive answer. )

Numerous times I have heard conversations regarding how high our bullets fly on their way to the targets. Some shooters opined that many of the flags that are 10 to 12 feet high were useless since the bullets reach 20 to 30 feet altitude. But looking at our rifles when we shoot, even at 1000 yards, I could not imagine any bullet going 30 feet high since the rifles are essentially level. I believe the error is that they assume that the peak of the trajectory is the same as the bullet drop at the target, and I believe this is incorrect.

Using the Berger Ballistic Calculator, I used 0 as the starting range, and 1000 as the ending range, and plotted the results for 180 gr Hybrid Target bullets at 2825 fps muzzle speed. These show the peak height is at around 500 yards down range, and is closer to 89 inches, or 7-1/2 feet:

View attachment 1479967

What do you think?

Thanks!
Alex

I don't see a calulatoer on the Berger website that produces a graph like yours.

My running data thru the Berger calculater.

All this means is (for example: The bullet drops 279" at 1000 yrds with a 100 yrd zero. If you want to be zeroed at 1000 yards you aim 279" above the intended poi. The scope is adjusted to aim 279" high but the cross hair is at the intended POI. Picture how high the rear sight has to be with iron sights to hit at 1000. For instance if you were GH hunting and the rifle was zeroed for 100 yards you would have to aim under the GH at any farther distance if the bullet increased in ht.

 

[6MMsteve]

The chart is misleading. The ht. as referenced to what. It's simple, draw a picture. Just remember the line of sight crosses the bullet trajectory. 26" high at 100 yards! Not sure what the chart represents. It isn't a drop chart with the gun zero'ed at the muzzle.

26 inches high at 100 got me dead on at Penn 1000yd club back in the eighties with a 308 Baer and 220 smk 86 grs 7828 ok

 

[Ned Ludd]

I don't see a calulatoer on the Berger website that produces a graph like yours.

My running data thru the Berger calculater.
View attachment 1480352
All this means is (for example: The bullet drops 279" at 1000 yrds with a 100 yrd zero. If you want to be zeroed at 1000 yards you aim 279" above the intended poi. The scope is adjusted to aim 279" high but the cross hair is at the intended POI. Picture how high the rear sight has to be with iron sights to hit at 1000. For instance if you were GH hunting and the rifle was zeroed for 100 yards you would have to aim under the GH at any farther distance if the bullet increased in ht.

The bullet never "drops 279" at 1000 yd with a 100 yd zero", not even close. That discrepancy is the entire basis for this thread. Your second sentence is correct, one would aim 279" high, or have the scope setting adjusted to aim 279" high in order to hit the intended POI at 1000 yd. But the bullet itself never drops anywhere close to that amount from its maximum ordinate during its flight to the target. The difference between the term "bullet drop" as it is used in the context of scope elevation setting, and true bullet drop as is defined by the maximum bullet rise over line of sight is significant and can create confusion.

In addition, for those claiming that the bullet can never rise above the the bore axis, that is not an entirely correct statement. It largely depends on how one defines the "bore axis". In fact, the barrel/bore does not represent a single perfectly straight tube, nor does its longitudinal axis represent only a single angle as the bullet traverses the bore during the firing process. If one defines the bore axis as a straight line, or even as the average bore axis over time, then in fact the bullet can rise above the bore axis after exit from the bore in a specific circumstance.

This circumstance occurs when a segment of the bore past the last inflection point (i.e. the last bending node), which can point both upwards and downwards during the barrel harmonic cycle, is pointing upward as the bullet exits the bore. This is how the process of positive compensation occurs, and it is easily demonstrated using a ladder test. In fact it is possible to have the muzzle end portion of the bore pointing upwards relative to the average bore axis angle, thereby increasing bullet launch angle to greater degree than what would be considered to be the average bore axis angle. Some may consider that description semanitics, but in fact it is not. For the purpose of the OP's original statement and question with respect to wind flag height, it probably doesn't matter as the effect is relatively small. Nonetheless, the barrel/bore is not a completely static structure during the firing process and that plays an important role external ballistics and the load development process.

 

[Conaso]

You know what, you are both correct! The ambiguity of the question is what allows this to be the case.

These are two more precise questions that will illustrate:

- After the bullets leaves the muzzle, does it ever rise above the plane at which the muzzle is pointing?
The answer is No. This is the question Butch answered and he is correct.

- After the bullet leaves the muzzle, does it ever rise further above the ground than the height at which it leaves the muzzle?
The answer is Yes. This is the question marine52 answered and he is correct.

The key is to make the question explicit and specify 'rise relative to what?'.

Relative to the plane at which is fired - No - because of gravity
Relative to the ground - Yes - because of launch angle

Did I mention Eötvos effect, shooting due East?

 

[Alexander-M]

Webster wrote:
"I don't see a calulatoer on the Berger website that produces a graph like yours."

That is correct. I used Excel to plot the graph from the numbers generated by the ballistic calculator.

Alex

 

[marine52]

The chart is misleading. The ht. as referenced to what. It's simple, draw a picture. Just remember the line of sight crosses the bullet trajectory. 26" high at 100 yards! Not sure what the chart represents. It isn't a drop chart with the gun zero'ed at the muzzle.

The original chart posted by Alexander-M is very easy to understand and replicate if you understand ballistics. As he said, he set the zero distance to 1000 and found his trajectory peak height to be 7.5 feet.

 

[6MMsteve]

set your rifle dead on at 1000 then drop back to 500yds see how far under the target u have to aim...... mens hearts will fail

 

[Rat Rifle]

Webster wrote:
"I don't see a calulatoer on the Berger website that produces a graph like yours."

That is correct. I used Excel to plot the graph from the numbers generated by the ballistic calculator.

Alex

Alex, how did You win all those awards shown in Your Avatar Picture without knowing the answer to this question ???

 

[Webster]

(This has been discussed somewhat before, but I have not seen a definitive answer. )

Numerous times I have heard conversations regarding how high our bullets fly on their way to the targets. Some shooters opined that many of the flags that are 10 to 12 feet high were useless since the bullets reach 20 to 30 feet altitude. But looking at our rifles when we shoot, even at 1000 yards, I could not imagine any bullet going 30 feet high since the rifles are essentially level. I believe the error is that they assume that the peak of the trajectory is the same as the bullet drop at the target, and I believe this is incorrect.

Using the Berger Ballistic Calculator, I used 0 as the starting range, and 1000 as the ending range, and plotted the results for 180 gr Hybrid Target bullets at 2825 fps muzzle speed. These show the peak height is at around 500 yards down range, and is closer to 89 inches, or 7-1/2 feet:

View attachment 1479967

What do you think?

Thanks!
Alex

The original question was what is the maximum height of the trajectory. In other words if the rifle is sighted in at some distance: for instance 600 yards what is the maximum height above the line of sight at 100, 200, 300, 700 ect. The bullet does start downward as soon as it leaves the muzzle. The barrel is simply pointed upward a small amount to account for the drop to the point of aim.

 

[marine52]

Yes, Alex did get his answer. His bullet rises above the point of departure (muzzle tip) to 7.5 feet which is the trajectory max ord. He was pointing out the confusion of some shooters who reference drop from the imaginary boreline which will always be above the trajectory. This imaginary line is the source of confusion. If you focus on where the bullet actually goes (trajectory) the confusion ends.