Incline and decline target engagement

[elkhunter11]

Forget the flight time for 800 m.Only the flight time over the horizontal distance matters.Think of it this way,if you shoot straight up or straight down,there can be a huge difference in flight time dependent on the distance,yet there is no horizontal flight time.As such,there is no holdover required regardless of the straight line distance to the target,which is of course correct.That example should in itself prove that you only use the horizontal distance and flight time to calculate holdover.Does that make sense to you?

[switchsl]

I understand, but feel it is incorrect, the flight time is relevant. In the example the target was 800m away the holdover was compensated to 725m. The holdover for a 725m shot would be calculated using a horizontal distance, when in fact in the equation the true horizontal distance is 800m. Using the standard holdover for a 725m shot compensates only for the difference in elevation based on its time to fall at 725m, when infact it is dropping over an 800m distance. There has to be compensation for the extra time the bullet will be in flight to reach the actuall distance of 800m instead of the 725m fixed point of impact your point of aim is adjusted to hit. There is a difference of .068, seconds and 75m for the bullet to change its drop based on the 725m hold over on a 800m shot. Now this brings me to the other question I had based on two different answers from two different equations of the same triangle.

The difference in time is .068 seconds, velocity = gravity x time
this means 9.8mps x .068 = .666m
the difference should be .459m

That leaves a margin of error of .207m if I calculate my drop based on distance to the target or based on the time allowed to drop. Using the 9,806mps as a constant for gravity, and using the same calculated flight times on both sides of the equation, why do I come up with such a drastic difference in values? the only variable is that I used time as opposed to distance to get the final answer, where did I go wrong?

[Bushrat]

I may be out in left field and not quite sure what your after I'm thinking that the difference in angles between the line of sight through the scope and the line of bore have to the point of impact must intersect. Line of sight being above line of bore, to have a bullet impact dead on at a specific distance the line of sight and line of bore have to intersect for this to happen. Add shooting at angles up or down changes the interaction of trajectory in relation to these angles and I believe you have a variable that influences the calculations causing real point of impact to vary from the calculated point of impact. I think if you used just the line of bore the calculated trajectories would be right on. Thinkin the differences in impact points your finding between real and calculated might have something to do with this???

[elkhunter11]

The holdover for a 725m shot would be calculated using a horizontal distance, when in fact in the equation the true horizontal distance is 800m.

The true horizontal distance is only 725 meters.The 800 meters is the straight line distance that includes the vertical element as well.

[switchsl]

Bushrat... I understand what you are saying. The line of sight and the line of the bore are both variables yes... you would have to add the height of your scope above the bore to find the angle, but the theory is suggesting that the line of sight be adjusted to impact at the intersection of line of bore for two given distances. The numbers in my calculation are based on a fixed point, that being the angle A and intersections of adjacent side b and hypotenuse c.

elk- What is the vertical element?

[Bushrat]

Originally Posted by switchsl

Bushrat... I understand what you are saying. The line of sight and the line of the bore are both variables yes... you would have to add the height of your scope above the bore to find the angle, but the theory is suggesting that the line of sight be adjusted to impact at the intersection of line of bore for two given distances. The numbers in my calculation are based on a fixed point, that being the angle A and intersections of adjacent side b and hypotenuse c.

elk- What is the vertical element?

The line of sight and line of bore sighted to intersect at say 100 yds on the level will change as the angle you shoot up hill or down change. If you take that same rifle calibrated for that scenario and point it with the line of sight at 90 degrees at a right angle shooting straight up in the air as opposed to level, the line of bore will still intersect line of sight at 100 yds but the bullet will intersect that line slightly before it reaches 100 yds and will never drop back down across that line, in fact it will keep going away from the line of sight, or " go over backwards" so to speak. If you shot it straight down at 90 degrees from the level the the line of sight and line of bore still intersect at 100 yds but the bullet will also cross the line of sight slightly before 100 yds and then keep going away and never cross back through the line of sight as gravity doesen't affect the trajectory of the bullet in the same way as bullets fired on the level regardless of time of flight. Anything in between level and 90 degrees up or down suffers the same effect in varying degrees depending on the angle it is shot at. I'm sure there is a formula that can calculate this but it would have to figure in the scope height to bore and point of impact angles and the varying gravitational effects on trajectory depending on the angle the bullet is shot at. I'm over my head here.... Hell, I'm not sure we're even talking about the same thing

[elkhunter11]

elk- What is the vertical element?

If I understand your example correctly,you have a straight line to the target distance of 800 m,and a horizontal component of 725 m.Is that correct?If it is,the 800m is the resultant of both the horizontal and vertical components.in other words,it is the hypotenuse of the right triangle.

If the 725m is the horizontal component as I am assuming from your description,you would only use the time of flight for 725 m,not 800m.
As I said earlier,only the horizontal component is considered for the time of flight and for the holdover calculations.

Again the example of shooting straight up or down comes to mind to prove this.You can shoot straight up at a target 100 yards away or at a target 1000 yards away,and despite the huge difference in time of flight,the holdover will be zero in both cases.Again ,the reason is that there is no horizontal time of flight,the time of flight is all in the vertical direction,which is irrelevant.

Of course,as Bushrat is trying to explain,you must also take the line of sight into account,since the basic equation relates only to the bore axis.You must do further calculations to relate the line of sight to the bore axis.These calculations include the sight height and zero point.

[switchsl]

You still dont understand... To correct for up/downhill shooting with a cosine indicator you take the lazered yardage and multiple by the cosine. This number would be the distance that the bullet is affected by the force of gravity. After determining this distance you simply look on your chart at the corrected distance and dial the dope for the corrected distance. This is the correct way to “triangle” the shot if you were shooting a lazer, which has no flight time. For the rifleman this method does not take into account that the bullet, while only being affected by gravity over the corrected distance still has a longer time of flight to be affected by. IF YOU USE THE ABOVE METHOD YOU WILL BE LOW EVERYTIME.

the triangle is as follows
side c-hypotenuse-directline to target-800m
side a-adjacent to angle A the incline angle of 25
side b-opposite, elevation of target from bore, 338.095

the distance of travel effected by gravity 725 = cosA
the distance to the target is hypotenuse c = 800
at 1100 fps the bullet has .068 seconds additional travel time

cos25 x 800m = 725.046m distance of travel effected by gravity
800m / 1100mps = .727s flight time
.5 x 9.806G x (.727s x .727s) = 2.588m drop in flight at 800m


725.046m / 1100mps = .659s flight time
.5 x 9.806G x (.659s x .659s) = 2.129m drop in flight at 725.046m
2.588m - 2.129 = .459m difference in impact point

Shooting straight up or down does not change the point of aim because drop in this case would not exsist, instead you would accelerate or decelerate the bullet speed only.

cos of 90 = 0
vertical trajectory motion equations used are
velocity2= velocity1-GT
H = (velocity1)(T) - .5G(T)(T)
Its to hard to write the entire equation in sequence without proper mathmatical symbols, but, in the end the bullet straight up will travel 61734.693m and take 112.245s to get there.
This proves that gravity has an effect on the bullet but has nothing to do with the vertical travel along the horizontal axis .5(G)(T)(T). The height of your scope and line of sight in relation to the bore have nothing to do with the influence of gravity or the time of flight of the bullet and is not valid in this instance. The height of scope and line of sight does come into play when you determine your intersection of line of sight and bullet path, eg to zero your scope at 200m. The question at hand relates only to the actuall drop in the bullet path, related to where the bore is pointing. If your scope is 1 inch above the bore or 3 inches above, the point of aim is still adjusted to intersect at a given distance. The intersection at 725 is pretermined, the intersection at 800 is unknown because of the variable of time. For this purpose in my question we only consider from bore to target in a straight line and bore to holdover point in a straight line. I now need to further understand the use of motion equations and how to calculate ballistic trajectory instead of straight lines. With your backround... maybe you can post the equation and work it for me. I stand by my answer, you will always hit low using a standard cosine equation, untill you prove me wrong with numbers.

[firegod74]

You guys are making my brain hurt with all this. But I'm pretty sure that by using the goood old High school SOH CAH TO/A, If you are using direct line of site for an uphill shot That would be Cos=Adjacent/hypotenues. Hypotenues being line of site and cos the angle. Adjacent ends up being your horizontal to target not bullet drop, and in the end you are talking about the same thing.

[elkhunter11]

It's hard to picture without a sketch to verify the situation,but I will try again.Given 800 as the hypotenuse,and 725 as the adjacent side(my interpretation of your right triangle,correct me if I am wrong) it would seem that you want to use the time of flight for the bullet to travel 800 meters.This will not work,the result will be a point of impact that will be low by the amount the bullet drops due to gravity from 725m to 800m.Humor me and use a standard ballistic calculation with 25m intervals and check the difference between 725m and 800m.Is the result the difference that you are ending up with?

[switchsl]

yes we understand the equation the same... hypotenuse is 800m, and 725 the adjacent side. My numbers are fictitious as I said, because I dont have a ballistics program. The discusion boards for various ballistics programs and one vs the other such as exbal, jbm, shoot etc. in long range hunting and sniping forums suggests that lesser programs especially the free ones dont take into account the extra travel time. My tables index at 100m so thats not accurate enough. To accurattly see this effect you would need to enter distance of 1000 or greater.... ill run the equation again using the same variables but 2000m instead of the original 800.

2000m target distance up a 25 degree incline velocity 0f 1100fps
effective gravity distance... cos25 x 2000m = 1812.615m
flight time for gravity effected distance... 1812.615 / 1100 = 1.647
drop at 1812.615m... .5 x 9.806 x (1.647 x 1.647) = 13.299m

cos value of holdover for a 2000m shot at 25 incline is 13.299

now, if you dialed in to 1812m because that is the gravity effected distance of the flight path you should hit the target but... the formula assumes the distance to the target is 1812m based on the flight time (1.647 x 1.647), this is incorrect because the actuall straight line to the target is 2000m, thus a diference of 188m of travel beyond what the formula allowed for the effect of gravity.

the actual flight time... flight time... 2000m / 1100fps = 1.818
difference = 2000m actual flight time less calculated gravity effected distance flight time
difference = 1.818 - 1.647 = 0.171
distance of drop in .171s= .5 x G x (.171 x .171) = .143m

Now to pull my head out of my ass... the first equation example is incorrect. The theory is the same, but i used the wrong procedure to calculate the difference in impact point aka the margin of error. This also answers my question of why I had two different values when i worked the equation in a different sequence.

the correct formula is...
cos25 x 800m = 725.046m distance of travel effected by gravity
725.m/ 1100mps = .659
800m / 1100mps = .727s flight time
.727 - .659 = .o68s
.5 x g x (.068 x .068) = .022m

the correct difference of impact due to margin of error in flight time is .022m not .459 as I stated above. .459m is the difference of impact between 800m and 725m. I hope this clears something up, and this is the reason I was looking for help with the formulas, I made a foolish mistake. Regardless the principle is the same, added flight time means added drop. The formula uses time squared to get a vertical drop, therefore as time increases drop increases exponetially. The formula firegod uses is correct, but the adjacent side is a calculated correction aka the distance gravity pulls the bullet down, not the actuall flight path aka hypotenuse so they are not refering to the same thing. Because they refer to different straight line distances, that is the margin of error in the equation because travel speed is fixed, distance is fixed, leaving time the only changing variable. Elk you stated in your last post, the exact problem I was trying to get across.

[elkhunter11]

Now I do see where you are coming from.It appears that your calculations have exposed a small margin of error in the method used by most shooters to deal with angles.We do use a shortcut,because it is much simpler,and much more practical than doing a long drawn out calculation for each shooting situation.However,it really isn't that big of a deal to the overwhelming majority of shooters.Lets face it,an error of .022m at 800m,is likely smaller than the error due to different velocities due to varying temperatures,along with other factors such as altitude,and humidity.Even a headwind or tailwind will add to the error.It can be challenging to figure out the exact numbers,but in most shooting situations,the extra accuracy has no practical value in real world situations.

[switchsl]

In a hunting situation, no application, in a target shooting application, some relevance if you are beyond 1000m. With the added variables of temp humidy, elev. wind the error could be exagerated even more, there are alot of variables to consider, for instance artillery gunners shooting at 12km its very relevant but not all of us get to try that on for size. For myself, as an example the long range capabilities of low velocity projectiles, such as a 45-70 or an 1853 Enfield musket. The slow travel time and long range capability of these rifles leaves a big margin for error, not all shooters are using high powered modern rifles remember. My 50 cal flintlock with a roundball and muzzle velocity of about 1800fps is a good example of this, with a ball, the BC is approx. 0.071 for a 185gr weight. Im trying to learn to make this calculation without the aid of a computer, which I can then use to double check my computer, develop my own tables, or argue in a forum untill my face turns blue. I thought some of the long range ballistics talk, and claims of 700 yard white tail kills there would be a better response to the theory of ballistics, rather than taking a table, range finder, or computer programs word for it. Next when my wife stops yelling at me Ill post the calculation for true ballistics instead of the straight line equations. For now its off to the honey do list.

[sullijr]

Someone is out to lunch.Drop is due to gravity only,whether it is up or downhill it doesn't matter.Time of flight does not matter a hoot,horizontal distance only is what matters.Your shot will be high of point of aim both up and downhill

[elkhunter11]

Someone is out to lunch.Drop is due to gravity only,whether it is up or downhill it doesn't matter.

Your shot will be high of point of aim both up and downhill

Nobody has posted anything that disputes any of that.

Time of flight does not matter a hoot,horizontal distance only is what matters.

How do you calculate the effects of gravity,without knowing the time that gravity is acting on the bullet?After all,the units for gravity are 32ft/sec/sec.

You like myself likely use a simplified ballistics calculations or tables that don't ask for time of flight,however,somewhere along the way,the people that came up with the ballistics tables or calculations,did factor in time of flight.

[sullijr]

if you shoot a bullet parallel to the ground and drop a bullet from the height of the muzzle they will hit the ground at the same time, the flight time is the same, gravity doesn't change it is constant.Galilao dropped a big ball and a small one from the leaning tower and they hit the ground at the same time.

There is a mad engineerat work on this thread.

[Traps]

Switchsl, where you went wrong here with time of flight differences is you didn't take into account the x component of velocity when figuring out flight time. Because your working with two components of velocity and distance, x and y or horizontal and vertical, you need to treat them as components of true distance or velocity.

horizontal direction (without the subscripts):
a=0
v=v
x=vt

vertical direction (without the subscripts):
a=-g
v=v-gt
y=vt-0.5tt

Our knowns are a=0, v = v in the horizontal direction so here we take cos (25) * 1100 fps = 996.94 fps, and horizontal distance of cos (25) * 800 = 725.04. We then figure out time from both the horizontal components of velocity and distance from x=vt, so solving for t we have t = x/v = 725.04/996.94 = 0.727 sec. The drop can be figured out then y = 0*0.727 - 0.5 * 9.81 * 0.727 * 0.727 = 2.59 m.

[Traps]

To state what might not be obvious is there is no difference in the point of impact, you have to set up your equations properly and you will come up with one answer, thats the right answer. Elk, I am disappointed, you should know better.

[elkhunter11]

if you shoot a bullet parallel to the ground and drop a bullet from the height of the muzzle they will hit the ground at the same time, the flight time is the same, gravity doesn't change it is constant.Galilao dropped a big ball and a small one from the leaning tower and they hit the ground at the same time.

If you shoot a bullet to hit a target 800meters away,you most likely aren't shooting it parallel to the ground.In fact it would be rare that you would shoot exactly parallel to the ground at any distance.This is especially true when you consider that our sights are normally above the bore axis.

Elk, I am disappointed, you should know better.

Traps,I haven't used those equations in many years,and I certainly wasn't about to dig into my books to rehash it all again.I am getting to old and too lazy to do that these days.As such,I never looked deep enough into his work, overlooked the error,and accepted that he might have found something that I just didn't remember.These days,I do forget some of what I learned many years ago.

[switchsl]

The error I made should have been obvious, like I said Im trying to teach myself physics from the internet based on what I remember from high school years ago. When i worked the question in two different ways there where two different answers, so there had to be a problem i just needed to identify it. The post was good, and what I was looking for... what you describe as Im starting to understand it is to seperate the vertical and horizontal component, and then, you can use the true ballistics calculations opposed to all the theory of straight line stuff like I started with. Im getting onto some of the steps now. Wish me luck...

[Traps]

Switchsl your tenacity is commendable. Let me know how it works out and if you need a hand with anything I might be able to help you out.

[switchsl]

Exellent traps I will take you up on that.