Slope/Angle ballistics (again!)

[Dave King]

Yup, me again.

Here's my latest puzzle (whether real or imagined it's still a puzzle).


Problem:
(Assume NO WIND)

Shoot a round at 45 degrees for a distance of 200 yards (this is the actual flight distance of the bullet).
The "gravity" or "base" distance is 141 yards so we correctly (or incorrectly) adjust the scope for 141 yards.
We fire the shot and it hits LOW.

My particular puzzle at this moment has to do with why the round hits LOW.
I KNOW that the bullet needs to travel 200 yards but it's on a 45 degree angle so I must adjust for the slope.

However, the bullet must also loose velocity over this 200 yard flight and it take time to make this 200 yards flight.
If I just assume that the 141 yards setting for the "gravity" or "Base" distance is the correct scope setting I cheat myself because it takes less time for the bullet to go 141 yards and it retaines more velocity over the shorter 141 yard flight. I cheat myself into thinking that a bullet traveling 200 yards on a slope can do it in the same amount of time it'd take to go 141 yards on a level trajectory. My bullet is traveling .05 to .08 seconds longer and will have an arrival velocity as much as 100 fps slower the the level 141 yard shot.

WHEW!!! Glad that out in the open...


Now, all you rocket scientists help me figure out how to account for this additional drop due to the slower velocity/added Time Of Flight (TOF).

I thought I might just figure the additional drop as if the bullet just left the muzzle and traveled the "additional" TOF, adding this drop onto the 141 yard drop data (it'd be somewhere in the .5 MOA area I guess).

I also thought I might "adjust" the BC to allow the bullet to make the 141 yard trip in the 200 yard TOF specification and use the drop for this "adjusted" bullet. (This doesn't sound like the correct approach but I'm grasping for straws here.)

Anyone still listening???

Thanks in advance..

 

[varminter]

I'm woking on it for you. But I need to draw some pictures and having problems gettin them to show up here. Not to computer savy mind you.
I try writting you out how I do it (meaning angle calc.) but it was way long and half confusing even for me and I wrote it. A picture says a 1000 words . It sounded a lot more complicated writing it out than it really was.
Hopefully be back tomorrow.

[ 02-20-2002: Message edited by: shootinlong ]

 

[Dave King]

I was just now falling asleep and it came to me. I also believe I've read or was told this method at one time also.

I use the cosine of the angle against the overall correction for the entire "slope" distance rather than calculating the "base" distance and using the drop for that shorter "base" distance.

For example:

200 yards total drop for .308 175SMK BC .507 MV of 2750. Total drop 10.1 inches, drop beyond 100 yard zero 3.5 inches, correction (scope) 1.66 MOA. 45 degree angle (cosine .707) correction amount = 10.1 * .707. Correction required is 7.14 inches for the 200 yards. About the correction required for a 170 yard "flat land" shot, 1 MOA over my "flat land" 100 yard zero.

This method invalidates the 100 yard zero as it isn't valid for an angled shot either. The 100 yard zero would be now placing the round .7 inches high and would have effectively become a 141 yard zero.

Hope this is correct, maybe I'm off in left field again.

[ 02-20-2002: Message edited by: Dave King ]

 

[Darryl Cassel]

Dave

If you use a longer distance to look at the difference of the drops it may become more clear if your trying to find out what I think your trying to find out.

Figuring that the bullet will shoot HIGH with a 45 degree angled shot, use 1000 yards and 750 yards as your base distance.

At 1000 yds with a 45 degree shot and using the 338gr 250 gr Bullet at 3250 FPS you will need 12.5 MOA from a 100 yd zero. The bullet path will be -131.26". Now lower that 1000 yards down to a 0 degree or flat straight away shot and you would need 18.5 MOA or
-93.23". A difference of 6 MOA just by flattening the angle of the shot.

Now go to 750 Yards and do the same thing. At 45 degrees you will need 7.4 MOA
(-57.94") and at 0 degrees of flat land shooting you will need 11.2 MOA(88.13")

Now comparing the 1000 yard 45 degree shot with the 750 yard 0 degree or flat land shot, you would have to take off ONLY 1.3 MOA to shoot at the 750 yard target.

The only reason I went with the 1000 yard and the 750 is because of the comparison of your 200 to your 141 yard shot.

The gravitational let up on a 45 degree shot is more then some think.

I'm not clear on what you said about being low with your 141 yard shot if your shooting at something 200 yards away and sighted in at that 200 yard target and on a 45 degree angle to start with???

I just ran the same bullet from above at 200 yards at 45 degrees and the 141 Yard shot at 0 degrees. At 200 yards with the 45 degree shot, if the rifle had a 100 yard zero (on flat ground), you would be on at the 200 yard mark (WITHOUT ANY ELEVATION CHANGE) also because of the lessor amount of gravitational pull.
Now go back to the 141 yard mark on a flat ground or 0 degree and your bullet path will be ONLY -.02" low or the SAME MOA as the 200 yard 45 degree shot was. That means it will shoot the same at the two distances.

The main factor is the 45 degree shot and I'm still not sure if that was your original question. The flight difference between a 200 yard and 141 or 150 yards when one is 45 degrees and the other is 0 degrees is not very much. It will show up a little bit more when you take the yardage out further but will still be close because of the 45 degree shot.

This is what my ballistics program just told me but, I'm still not sure if that's what you were after????

My question is, how many 45 degree shots would you have at 200 yards and then come back to 0 degrees for a 141 yard shot? If you were shooting 45 degrees and only 200 yards distance, I doubt if there would be an animal able to stand on that steep hill?

Darryl

[ 02-21-2002: Message edited by: Darryl Cassel ]

 

[varminter]

I finally got the image in. A little to late though I see. Your second post is right as far as I know. Or should I say that's the way I do it anyhow. I do alot of angle shooting in Tioga county on varmints and eastern Montana on mule deer and whitetails. And this method seems to work for me.
I included the picture because it's an illistation of what you said and because I finally figured out how to do it.
Sorry I wasn't more helpful.

[ 02-22-2002: Message edited by: shootinlong ]

 

[Brent]

Dave,
I assume your problem is a result of using a GPS that gives you a flat land, or a "base" distance only unlike a rangefinder does.

If I understand you correctly, you are atempting to find the moa setting for a zero at 200yds "line of sight" (LOS) at a 45° angle.

Basing this on the "base" 141yd distance or the LOS 200yd distance would give an incorrect moa setting.

The 141yd setting (only the GPS gives) would, yes cause a below LOS impact.

The 200yd setting (the laser RF gives) would, cause a classic above LOS impact.

What you want to do is convert the base distance to actual (laser) distance so you can then correct for the 45° angle, which the later part the Oehler Ballistic Explorer program does, true?

And you want to do this using known data from a GPS and something that measures the angle for you, without having to ever know the actual LOS distance, correct?

But, I'm probably wrong because you used a 200yd range in your last post so should I assume you're using a rangefinder now also?

I admit, I am a little confused about why you were trying to incorperate the "base" distance into this, then abandoned it in the end? Were you simply trying to figure the 45° angle into the "come ups"?

I also came up with 10.6" total drop and 1.81 MOA correction at 200yds on both programs I have using your specs at standard pressure. Also the BC "Load From A Disk" gives is .496 for that speed.

BTW, what do you use for measuring the angle and where do you buy one? Also how do you figure the "cosine" for the angle as you did here (.707)? I looked up cosine in the dictionary an it said I had to know trigonometry to figure it, which is obvious now that I don't. Feel like giving a lesson? If so go SLOW.

Dave, what does the astrisk signify (*) in the "=10.1 * .707" you wrote? is this another sign for multiplication? I assume it is as the "7.14 inches" answer you gave is arived at by this.

Sorry, I hope I haven't confused anyone. I only intended to clarify a few points as this is something I have been wondering about this last week but have not brought up yet, but Dave opened the door now.

You are the one who often mentions the slope angle or cosine in posts and has recently been nagging at me because I don't understand enough about how to calculate it's effect without the computer program or how to measure it while in the field. We will be hunting below some steep mountain sides this fall so it maybe necessary to figure in.

Later
Brent

 

[Dave King]

Dave,
I assume your problem is a result of using a GPS that gives you a flat land, or a "base" distance only unlike a rangefinder does.
If I understand you correctly, you are atempting to find the moa setting for a zero at 200yds "line of sight" (LOS) at a 45° angle.

This is exactly correct.

Basing this on the "base" 141yd distance or the LOS 200yd distance would give an incorrect moa setting.

Again, this is correct.

The 141yd setting (only the GPS gives) would, yes cause a below LOS impact.

The 200yd setting (the laser RF gives) would, cause a classic above LOS impact.


Correct again.

What you want to do is convert the base distance to actual (laser) distance so you can then correct for the 45° angle, which the later part the Oehler Ballistic Explorer program does, true?

I'm a little fuzzy here. I don't have the Oehler Ballistics Explorer software so I'm unfamiliar with it's functioning.

By using the 141 yard "base" distance in conjunction with my elevation readings from the GPS I KNOW the "slope" or distance that a LASER RF would indicate. It's the Pythagorean theorem (A squared + B squared = C squared) thing.

Remember that I have both the target's GPS fix and the shooter's GPS fix, these fixes also give me elevation. I get the height of the triangle leg by the difference in the two elevations and the length of the base from the difference in flat land distance.

And you want to do this using known data from a GPS and something that measures the angle for you, without having to ever know the actual LOS distance, correct?

The computer computes the angle from the two sets of data, 2 postions and 2 geographic elevations. I have the height and base lengths. The height and base angle is always 90 degrees because the center of the earth is directly below us.


But, I'm probably wrong because you used a 200yd range in your last post so should I assume you're using a rangefinder now also?

I'm sure you've got it figured out now but I state it anyway. No I still haven't used a LASER RF.


I admit, I am a little confused about why you were trying to incorperate the "base" distance into this, then abandoned it in the end? Were you simply trying to figure the 45° angle into the "come ups"?

The "base" distance is the DISTANCE gravity will act on the bullet.

I also came up with 10.6" total drop and 1.81 MOA correction at 200yds on both programs I have using your specs at standard pressure. Also the BC "Load From A Disk" gives is .496 for that speed.

BTW, what do you use for measuring the angle and where do you buy one?

With two GPS fixes and the elevation at those fixed points you can calculate the angle with a pocket calculator.

Also how do you figure the "cosine" for the angle as you did here (.707)? I looked up cosine in the dictionary an it said I had to know trigonometry to figure it, which is obvious now that I don't. Feel like giving a lesson? If so go SLOW.

I'm pretty good at going slow, I spend so much time tripping over the obvious that I barely make any headway ever.

Here's a cosine table and a link (the link is a PDF).

angle cosine
10 .9848
15 .9659
20 .9397
25 .9063
30 .8660
35 .8192
40 .7660
45 .7071
50 .6429
55 .5736
60 .5
65 .4226
70 .3420
75 .2588
80 .1736
85 .0872

http://api.ctcd.cc.tx.us/consume/a104gm8.PDF


Dave, what does the astrisk signify (*) in the "=10.1 * .707" you wrote? is this another sign for multiplication? I assume it is as the "7.14 inches" answer you gave is arived at by this.

Yes, it the sign for multiplication.

Sorry, I hope I haven't confused anyone. I only intended to clarify a few points as this is something I have been wondering about this last week but have not brought up yet, but Dave opened the door now.

You are the one who often mentions the slope angle or cosine in posts and has recently been nagging at me because I don't understand enough about how to calculate it's effect without the computer program or how to measure it while in the field. We will be hunting below some steep mountain sides this fall so it maybe necessary to figure in.

I'm always trying to understand how to better my chances at making a first round hit from calculated data.


You can make a little angle indicator for field use by using a plastic protractor, a piece of string and a weight. The MilDot Master has a provision for doing this too.

 

[Charles A]

This is extremly easy but just in case I'll put it laymen terms. Here's an example
You range a target at 1200yds.The slant angle is 30 deg. The cosine facter for 30 deg. is .8660.
Take the slope angle, .8660 times it by true range(1200yds) and you get 1039yds this is your slant range.This is how much the bullet will drop because of gravity.But rember your windage data will still be based on 1200yds because the bullet has to travel through 1200yds of air.
Any time you can lower your MOA setting your in a gain position. The nice thing about sloped shots is it gives you some leeway on your range estimation, it increases your danger space.It gives you more margin of error for first round hits.

[ 02-22-2002: Message edited by: SEALSNIPER ]

 

[Dave King]

SealSniper

This is extremly easy but just in case I'll put it laymen terms. Here's an example
You range a target at 1200yds.The slant angle is 30 deg. The cosine facter for 30 deg. is .8660.
Take the slope angle, .8660 times it by true range(1200yds) and you get 1039yds this is your slant range.This is how much the bullet will drop because of gravity.

Initially I thought this was all there was to it also but there's more to consider. We know the "gravity" distance is 1039 yards but the distance the bullet must travel is actually 1200 yards, an additional 161 yards.
This 161 yards takes some amount of time and as a result the bullet slows down due to drag. So even though the gravity distance is 1039 yards the bullet is in flight for the amount of time a 1200 yards shot takes. It's muzzle velocity is the same no matter the distance of the shot, the "terminal" velocity is different for the angled shot.

The amount of drop is more than the straight 1039 yards but less than the 1200 yards. The actual amount is still a bit of a mystery for me.


But rember your windage data will still be based on 1200yds because the bullet has to travel through 1200yds of air.

This is sometimes a forgotten consideration.

Any time you can lower your MOA setting your in a gain position. The nice thing about sloped shots is it gives you some leeway on your range estimation, it increases your danger space.It gives you more margin of error for first round hits.

Thinking about "danger space" is a good thing, I consider this on nearly all shots. It's a good bit different for a short vertical target as hunters see and also for a chest & head only for military. (Sometimes I'm quite sure there's a God as he proaects the animals by making them more horizonal then vertical.

)

 

[Charles A]

DAVE, I dont have the tables with me but I think I remember most of it.I didn't put this in at first because it gets a little complicated(for most).Theres three other things you have to account for -TEMP-BP-AMMO TEMP.First you will need a scientific calculater.
Where going to use my gun.
30cal 220gr BC .640 MV 2820 sighted in at 100yds in starderd condtions.
For BP if its up-multiple by constant,if down divide.
For temp if its up divide by constant,if down multiply.
Here's the shot 1200yds,30 deg. angle,BP 27.55,temp 84 deg.,ammo temp 84 deg.
First we find the slant range 30cos*1200yds=1039
Then the diff. in BP. 1.980. the constant for 1200yds is 1.0367 (I think)
enter 37.4 divide by 1.0367 hit "xy" key enter 1.980(the diff. in BP) hit =34.82398 this goes where BP is on the table.

find the diff. in temp. 25deg. the constant for 1200y is 1.03426
enter 37.4 hit divide enter 1.03426 hit "xy" key enter the diff. in temp (here you move the the decimal point to the left) 2.5 hit =34.37931 this goes in temp on table.

find the diff. in ammo temp. 14deg. the general rule of thumb is +/- 1MOA for every 50fps. diff. in temp. where going to use -.25MOA this goes in the ammo temp table.

Remember that you use the TRUE RANGE for all calculations.After you get EACH new MOA elevation setting subtract EACH one of them from your intial setting,and put them in the space labled DIFF.

INT.MOA SETTING DIFF.
TRUE RANGE 1200y 37.4MOA
SLANT RANGE 1039y 29.3MOA
BP 27.55 34.82398MOA -2.57602MOA
TEMP 84 34.37931MOA -3.02069MOA
AMMO TEMP 84 25deg change -.25MOA

Add all the differences up and subtract or add them to the SLANT RANGE moa setting.Here it comes to 23.45329MOA THIS THE REAL SETTING YOU PUT ON THE SCOPE.This is the setting you would use.

WHEW

I know this was kind of long and its not exact because I dont have the tables with me,but its close.The bullet does have to fly through 1200yds. of air.This is how we account for it.
For everybody that wants first round hits at longrange,past 800yds or so YOU MUST MAKE THESE CORRECTIONS!!!!!

SINCERLY THE GRIM REAPER...

[ 02-23-2002: Message edited by: SEALSNIPER ]

[ 02-24-2002: Message edited by: SEALSNIPER ]

 

[Brent]

Thanks Dave,
I have a few more questions but later.

Sealsniper,
You came up with 23.5 MOA corrected for 1200yds?

My programs, both the Oehler Ballistic Explorer and Load From A Disk indicate a larger total drop of 29.3 MOA, and 30.37 MOA respectivly at 1200y with your conditions entered.

At standard conditions with 30° angle also are 32.5 MOA.

1039y drop was 28.1 MOA and 27.2 MOA at your conditions on flat land which does not equal the 23.5 MOA. As Dave suggested it falls somewhere in between the base distance and actual one.

I have some questions about the constants for BP and temp etc but I have to run now.

Later
Brent

[ 02-23-2002: Message edited by: Brent ]

 

[Charles A]

Brent,I dont guess you caught it in the last post but I said that I didn't have the tables with me.Your right,the right setting would be 27.9MOA,as per JBN ballistics.Thats with out accounting for ammo temp.Putting the ammo temp in it should be 27.75MOA.(I origenily said .25 MOA for ammo temp but it will be more like .8MOA).
My spotter has the tables with him,although I should get them back by the end of the week.When I do I will put what it should be,acording to them.
As for the bullet having a longer TOF. between the two ranges,yes but we have used these tables out to 1100yds and they have worked,even on sloped shots.I've been trying to figure it out but cant so I will ask and see if a few people I know can help.

 

[varminter]

Here are some formulas that may help out.

Drop = -1/2g*t2 or drop is equal to one half constant of gravity times time squared.

So in slope fire we are looking for the component of gravity at said angle.

slope drop = -1/2g*cos Angle * t2 or one half constant of gravity times the cosine of the slant angle times time squared.

The easy way to figure this is to take the drop for level fire at said range and multiply by the cosine of the slope angle. This will give the drop for slope fire.

We are assuming time of flight is the same so no problems there. We are using the actual range so we don't have problems with energy and remaining velocity either.

I hope this helped a little.

Greg
ps.
If you use the formula above for drop you will get a little more drop than shown in most tables and programs especially at longer ranges.
To keep it simple I didn't include the drag component which at longer ranges the drag vector tips up and acts against gravity. This function is built into most tables and programs so I just use the drop inches at range multiplied by the cosine of the slope angle.
Greg

[ 02-27-2002: Message edited by: shootinlong ]

 

[Dave King]

Greg

"The easy way to figure this is to take the drop for level fire at said range and multiply by the cosine of the slope angle. This will give the drop for slope fire."

This is what I assumed was correct after giving the problem some thought.

Thanks for the reassurance.

/r