Incline and decline target engagement

[switchsl]

I have read some good discusion on this forum regarding long range shooting and I was wondering if anybody could clarify the subject of riflemans rule and the practical application of hitting your mark on uphill and downhill targets. Im not a skilled, experienced or for that matter competant marksman but I do take interest in the sport and especially in its theory. I have fired hundreds of rounds in competition at various locations up and down hill, and can say I understand that it changes your point of aim at a given distance. What I cant get my head around is the physics involved, and certianly the mathmatical equations relating to it. I have read several articles on the subject but my comprehension of the math is limited. Any simplification, or direction to literature online I may not have found yet would be much appreciated.

[elkhunter11]

http://www.loadammo.com/Topics/April04.htm

[switchsl]

good article elkhunter11, thanks. I think what I really need is a high school refresher on my trig functions. The basic formula as I read it is cos x incline angle x actuall distance = the compensated distance

eg a 600 yard shot on a 30 degree incline gives you an actuall aim point for 519.6 yards

the equation is as follows cos(30)x600=519.6

this can then be used with trajectory tables to determine the corect holdover point.
It comes up alot in archery discusion, because of the elevation from tree stands.

[switchsl]

These are Some of the pages I found usefull.
http://www.exteriorballistics.com/eb...ned/5th/33.cfm
http://www.longrangehunting.com/foru...68/index2.html
http://library.thinkquest.org/20991/alg2/trig.html

Or buy a rangefinder with a built-in ballistic range calculator....

You need a means of accurately measuring the distance and calculating the angle regardless. Having one unit that does it all seems the simplest.

[elkhunter11]

The truth is,that outside of abnormally steep angles,or extremely long ranges,if you are using a relatively flat shooting centerfire rifle,the shot angle is seldom an issue.For archery hunting out of treestands,it can be a concern.

Originally Posted by elkhunter11

The truth is,that outside of abnormally steep angles,or extremely long ranges,if you are using a relatively flat shooting centerfire rifle,the shot angle is seldom an issue.For archery hunting out of treestands,it can be a concern.

If you actually run the numbers, it's not much of a concern in treestand hunting with normal treestand heights and ranges. There is much advertising hype there. As you pointed out, long ranges and steep angles is where it really comes into play.

[elkhunter11]

If you actually run the numbers, it's not much of a concern in treestand hunting with normal treestand heights and ranges. There is much advertising hype there. As you pointed out, long ranges and steep angles is where it really comes into play.

Agreed,under most conditions,the shot angle is not a significant factor at all.Like many other things,it is overblown by the manufacturers that want to convince us that we absolutely must have their products to be a successful hunter and shooter.

[switchsl]

Yes, I could buy lots of gizmos so I could just dial and shoot. Im just interested in how to make the caculations myself. I guess I have to much time on my hands. I like to know why things work, and how you come to the conclusion that it does work. Now I realise real world conditions change the exact numbers from the Equations, and Im not interested in what practical application the numbers have to real world situations, thats a debate that can go on for days. The entire projectile physics and its application to exterior rifle ballistics is just a curious subject for myself.
One thing I am having a hard time to gasp is why cos(angle) x holdover distance= corrected holdover?
What I guess i cant grasp is why cos(angle) x holdover will equal corrected holdover. can somebody explain why cosA gives me this number?

Originally Posted by switchsl

Yes, I could buy lots of gizmos so I could just dial and shoot. Im just interested in how to make the caculations myself. I guess I have to much time on my hands.

Regardless of how you do it, you will need gizmos. A laser range finder is a must as is something for measuring the angle. There's no way around it.

[elkhunter11]

If you want to know the exact distance,and the exact holdover and windage for a theoretical perfect impact,you need a good rangefinder,and accurate angle measuring device,a chronograph,a thermometer,a barometer,a humidity measuring instrument,and wind measuring equipment that can measure the wind strength and direction all the way to the target.Even if you know the exact distance,the velocity of the bullet will change with varying climactic conditions.The wind will vary even more.

As for myself,I use a laser rangefinder built into my binoculars to estimate range,I estimate angles when I think that they will be a factor,and I estimate the windage.I am not going to take the time and effort to complicate things too much in an effort to make things exact.In fact in many situations,it wouldn't be practical to even try to make things exact.

As it is,using the gear and the system that I use,I haven't missed an animal as a result of not knowing the exact distance to the target.I have however misplaced a few shots because I misjudged the wind.

[switchsl]

I know I need the gizmos... thats outside the scope of this thread. Its about the numbers and calculations... not the gizmos that give you the numbers to make the calculation. I want to learn to aquire the final answer to the question and understand how I got there. For now hypothetical numbers on a page are sufficient for that purpose. When I want to apply this knowledge, then Ill have a high tech something or other in some breed or flavour of my choice to give me the values to apply it... untill then, Ill learn to walk before I run and discuss the newest running shoe to do it with.
I want to learn about algebra, and projectile motion not the latest in sales tactics and electronic shooting aids.

This is an interesting site I also found relating to the subject
http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html

[elkhunter11]

I want to learn about algebra, and projectile motion not the latest in sales tactics and electronic shooting aids.

The simplest explanation,is to use the horizontal distance to the target.That is the effective distance that gravity acts on the bullet,and that is all that the gizmos are used to determine.

Originally Posted by elkhunter11

The simplest explanation,is to use the horizontal distance to the target.That is the effective distance that gravity acts on the bullet,and that is all that the gizmos are used to determine.

Yup!

[switchsl]

If you use the actual distance to a target to calculate the actuall horizontal distance that gravity effects the projectile you will consistently come up with a value that will put your point of impact lower than your calculated point of aim.

If you use the actuall distance to the target and calculate the holdover based on the known trajectory of the projectile, then from that information calculate your point of aim you will be consistently closer to the actual point of impact.

The subject is not that simple, and its my fault for saying I was looking for a simple explination... I am looking for a mathmatical explination for this, examples of those equations.

Originally Posted by switchsl

If you use the actual distance to a target to calculate the actuall horizontal distance that gravity effects the projectile you will consistently come up with a value that will put your point of impact lower than your calculated point of aim.

If you use the actuall distance to the target and calculate the holdover based on the known trajectory of the projectile, then from that information calculate your point of aim you will be consistently closer to the actual point of impact.

The subject is not that simple, and its my fault for saying I was looking for a simple explination... I am looking for a mathmatical explination for this, examples of those equations.

No, it really is that simple. The true horizontal distance is the true ballistic distance or at least close enough in a hunting situation. The actual distance to the target is somewhat irrelevant unless you are using it in your calculation.

[elkhunter11]

If you use the actual distance to a target to calculate the actuall horizontal distance that gravity effects the projectile you will consistently come up with a value that will put your point of impact lower than your calculated point of aim.

If you use the actuall distance to the target and calculate the holdover based on the known trajectory of the projectile, then from that information calculate your point of aim you will be consistently closer to the actual point of impact.

If you use the actual horizontal distance to the target,then calculate the holdover based on the trajectory of the load,you will be as close as is possible.

[switchsl]

Thanks for your input. I dont think you and I are on the same page here. Close enough and what situation like I said is irrelevant. Im interested in the scientific calculation... only for the purpose of learning to make those calculations. The distance to the target is always relevant, its one of the variables in the equations that in the application of target shooting is always rellevant and needed to make any further calculations. To discourage further argument, lets leave it at that unless you can explain to me... mathmatically... not in your experience, or hunting situation the answer to my question.
Why does the cos(A) value of distance give a less accurate value than cos(A) of holdover.

[fordtruckin]

I know I didn't ask the question but thanks for posting that link elkhunter. I have always wondered about the calculation for elevation but never thought about it untill this post. That is good stuff!

Going into trig here.. With a right triangle a2 + b2 = c2, a is your run, b is your rise, and c is the hypotenuse.

the square root of "c" is your line of sight or the hypotenuse

So: "a" is your run, in this case true distance, "b" is your rise, or in this case, total elevation gained or lost depending if your shooting uphill/downhill

CosA = a / c

"a" being run or true distance and "c" being line of sight

a little algebra to get "c" on the other side of the equation gets you:

"CosA x c = a"

"A" being the angle your shooting at up or down, "c" being the line of sight, and "a" being your true distance or the distance you would use to figure out the true distance to your target.

Its all math, it took me about 45 minutes trying to remember how it works... hope this helps

[elkhunter11]

To put things at their simplest:

Determine the true horizontal distance to the target-for example 300 yards

Determine the holdover required for that true horizontal distance to the target.-300 yards

Use that holdover and shoot.

Forget the straight line distance to the target,and the holdover for that distance,they are not required at all.

Originally Posted by elkhunter11

To put things at their simplest:

Determine the true horizontal distance to the target-for example 300 yards

Determine the holdover required for that true horizontal distance to the target.-300 yards

Use that holdover and shoot.

Forget the straight line distance to the target,and the holdover for that distance,they are not required at all.

Yup!

[switchsl]

Thanks ford. Yes it does clarify my question in some respects. I think I should have read further into the subject before I posted a thread on it. As I read and get involved in more obscure variables like the effects of gravity, time of flight, and so on, Im begining to understand and remember some of these formulas and how to work them in different order. Its been a long time since I was in school for sure, and schooling is what I was asking for. I thought possibly some of the military guys trained in artillery maybe able to shed light... just incase they have a howitzer season on chernobyl varmints.

[switchsl]

Once again and hopefully for the last time… Im interested in making the calculations for this problem. That’s what is simple about it. There is more than one way to make the calculation, depending on the trig function you use, and the variables you plug in. Its entirely incorrect to say that any variable be forgotten.
Unless you can post some usefull information and equations please stop cluttering the thread and perhaps start educating yourself beyond the first google’d response you come across.

[elkhunter11]

Thanks ford. Yes it does clarify my question in some respects. I think I should have read further into the subject before I posted a thread on it. As I read and get involved in more obscure variables like the effects of gravity, time of flight, and so on, Im begining to understand and remember some of these formulas and how to work them in different order. Its been a long time since I was in school for sure, and schooling is what I was asking for.

I have a background in mechanical engineering,and I have worked a great deal with the equations and calculations that are involved,however,this is really so simple,that a person need not bother to look so deeply into the calculations,or how they are derived.It really is as simple as using the true horizontal distance,and the holdover for that distance.To do otherwise,simply makes things more complicated than they need be to determine the proper holdover.

Unless you can post some usefull information and equations please stop cluttering the thread and perhaps start educating yourself beyond the first google’d response you come across.

I used the link as an example,because it was quick to find,and it serves the purpose.As for my education,my background is in Mechanical Engineering,and I earned my qualifications by spending years in school learning statics,dynamics,thermodynamics,applied mechanics etc.I just don't see the point in over complicating things when it serves no practical purpose.

[fordtruckin]

No problem switchesl.. Everyone here doesn't really seem to get that you don't care about using it.what you care about is knowing how and why it works. Sometimes it is just more information for yourself...

As for all those variables.. yeah it would be impractical to factor all those in for a 300 yard shot on a deer.. If you were shootin 1000+ yards then I would take a greater look into.. but then at 1000 yards the wind, temp, all that stuff is going to change from where you took the shot. Good luck!

[switchsl]

Originally Posted by fordtruckin

No problem switchesl.. Everyone here doesn't really seem to get that you don't care about using it.what you care about is knowing how and why it works. Sometimes it is just more information for yourself...

YES!. I know there are ballistics programs that can do this for me... just like I can use an autocad program to make a drawing instead of a pencil. I want to understand the mathmatical process in all respects to finding the values the program calculates for me. Simply calculating corrected distance and getting a holdover for it is not good enough. Most values are not relevant untill you are at least 1000m away, 2000 even... but thats what I want to determine for myself, and be able to use the values to check and recheck my work.

[elkhunter11]

If you really want the long drawn out equations,here is a link that should help you.Again,it is one that I could quickly find.Have fun.

http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html

[switchsl]

I already posted the link, and I have yet to see an example of your understanding of the long drawn out equations, or even a clear response to any question i have proposed. Elkhunter, please keep your coments to yourself if you are unable to read and evaluate previous posts. I made it clear I was looking for examples of the question I posted above. I have already answered my own question through research. Thanks for the input and demonstration of your complete lack of respect. I can only assume you feel I am not worth your time to teach, you are incapable of decent interaction, or you are a wanna be based on other irrelevant info posted in so many other threads.

I appologise for posting this question in the first place, and regret not being able to retract the entire thread.

[elkhunter11]

I can only assume you feel I am not worth your time to teach, you are incapable of decent interaction, or you are a wanna be based on other irrelevant info posted in so many other threads.

I was trying to help,as was demonstrated by me posting the original link,but based on your last post,I will go with the first part of the quote.

[switchsl]

I lost my cool, and I shouldnt have taken it out on you elkhunter, lets agree to disagree for the moment. Ignorance aside, help me understand the problem in this calculation. As an engineering student Im sure it should be straight forward. Im here to learn, and debate in a productive manner, and I see a flaw in your statement that the calculation is close enough.

fictitious numbers for calculation sake... 800m 25 incline 1100mps

cos25 x 800m = 725.046m corrected holdover distance
800m / 1100mps = .727s flight time
.5 x 9.806G x (.727s x .727s) = 2.588m drop in flight
725.046m / 1100mps = .659s flight time
.5 x 9.806G x (.659s x .659s) = 2.129m drop in flight
2.588m - 2.129 = .459m difference in impact point

If I was to take my theoretical distance calculated by correcting the 800m with cos of the incline I would in theory miss my target by .459m. This is not close enough to assume I can leave anything out of the calculation. Now, I realise this is a projection of straight lines, not the parrabellum of a true bullet flight, but assuming temp. humidity wind and barom. pressure are all the same it leaves a margin of error to great to be ignored. I understand the variation of decreased excelleration over time, but again assuming the same variables acting against the bullet decelleration of the bullet by gravity on a 25 incline is negligible, amounting to only fractions of a cm.