Go | New | Find | Notify | Tools | Reply |
<Don G> |
Big John, The answer is in the barrel vibrations. Your bullet must be leaving the barrel when the muzzle has an upward velecity. The bullet leaves the barrel with this extra upward component that cancels some of the drop due to gravity. A crude estimate says that the upward component must be about one foot per second. That is an entirely believable number. I would write down the load and store it away very carefully. It is a good one for your rifle! Don | ||
one of us |
The Oehler Ballistic Explorer software we use will describe the path of a bullet if you give it a distance at which the rifle is zeroed. If you do not specify the zero, it gives the path as dropping from the muzzle out. It sounds like this is what you are getting. Your range results indicate that the first line of sight crossing is at 100 with the zero somewhere beyond 200 which does not quite make sense to me. Is the range on an incline? Remember that the bullet crosses the line of sight twice, once going up and then coming down. ------------------ | |||
|
<Big50> |
The bullet path you describe is zeroed at 210yds, also .9" high at 100yds, .3" at 200yds and 3.9" low at 300yds or it is zeroed at 285yds and is 2" high at 100yds, 2.5" high at 200yds and .7" low at 300yds. It is zeroed at one or the other, it can't be both. Were the 2 and 300yd ranges single shots or groups measured at center? Something is WAY off and it's not the computer program, mine was the same also. Later | ||
<Harald> |
Go back to the range and in a single day shoot 100, 200 and 300 yard groups without adjusting the sights in any way. I think you'll find that the 100 zero does not match the reported 200 and 300 yard impact points. The BC also looks a little optimistic for a short bullet. Gerard is right though that the trajectory you described as an output does not take into consideration the zero range (it is set to the muzzle). Most scopes are assumed to be about 1.5 inches above the line of sight, so the bullet begins at a -1.5 inch position and climbs (due to the slight upward angle of the bore) to the line of sight, arcs upward and then downward in a parabolic trajectory, crosses again somewhere downrange (zero range) and continues to fall to earth. Fired perfectly flat it obviously always falls. | ||
<Don Krakenberger> |
Years ago I did a printout for a friend's 300 wby and somehow it magically seemed to shoot much flatter than the printout. I think I like the explanation about barrel vibrations. Could gun recoil as the bullet exits also affect this--or, at the risk of sounding really dumb could a barrel actually be mounted NOT parallel to the line of sight? | ||
one of us |
Harald, The length of the 130 gr HV is 1.31", about the same as a jacketed 160 gr spitser boat tail. Don, ------------------ | |||
|
one of us |
I'm guessing your range is on a hill (or different hills at different ranges) and/or the ranges to your targets are not accurately measured and/or something's loose in your rifle causing its zero to change. The stuff you posted above simply isn't physically possible. I've never seen any evidence of any "barrel vibration stuff" affecting the laws of physics. If the bullet is X inches high at 100, it doesn't give a damn what the barrel was doing way back when--it's simply flying (hopfully straight!). If your testing conditions are controlled and the computer modeling accurate, you'll be so close that the accuracy of your rifle (or lack thereof) will prevent you from detecting any computer errors. Second, you can't use a G5 curve with a BC meant for G1 curves. For example, a .550 BC based on the G1 curve is equivalent to a .309 BC if you want to use the G5 curve. And to all that find their rifles are flatter shooting, not as flat shooting, etc...as the computer says: Assuming your test conditions are completely accurate (level, measured ranges, rifle accurate enough to tell the difference, etc) are you measuring the atmosphere and making the necessary corrections? For example, I grew up at "only" 4400 ft altitude. Under "standard" conditions that means my bullets with a .550 BC have a corrected BC of .627. Now, if I shoot on a 90 degree day, the BC goes up to .695. That is easily enough to make a noticable difference. Garbage in = Garbage out. | |||
|
<nated> |
jon a, where did you get that info on bc, altitude changes, temperature effects on bc? that's very interesting to me. thanks, nate | ||
one of us |
nated, All that stuff is standard input required on Oehler Ballistic Explorer software. Very interesting and quite an eye opener. ------------------ | |||
|
<Harald> |
A few years ago I wrote my own ballistic computer to perform all these calculations. It uses a 4th order Runge-Kutta numerical integration of the set of coupled differential equations that result from the Siacci solution to the otherwise insolvable complete equations of motion (ie, assumes flat fire, neglects coriolis and magnus forces, etc.). I drew on the equations presented in the Sierra manual and published G1 drag data found in several basic sources. I have a 64-bit PC executable that I can email to anyone who wants it. It will allow you to correct for local weather and altitude conditions if you don't want to use the standard values. There is a terminal ballistics calculation module, but its not real - I just put it in as a placeholder to test the fuction of the main program. The problem with the barrel vibration theory is that any additional force imparted at the muzzle will alter the trajectory at all ranges, not just 100 yards. I haven't done an exhaustive analysis but I am inclined to agree with JonA that the trajectory you describe isn't possible. I don't have my calculator with me because my home computer is a Mac but just glancing at a table for BC .55 and 3500 fps I predict that you will get a trajectory height of 2.5 to 3 inches at 100 yards for 2.5 inches at 200 yards and that it would have a zero exceeding 300 yards. Anything truly zeroed at 100 yards would be -1.7 in at 200 yds and -7 in at 300 yds. If you were actually about 4 inches low at 300 yards then you would have to be near zero at 200 yards and almost an inch high at 100 yards. Something clearly is mixed up. I think you will find that with a high BC and very high velocity, it is a waste to have a zero range less than 250 yards and I would think seriously about 300+. My calculator will tell you the optimum zero range for a 4 inch maximum trajectory height. Sorry Gerard, for some reason I was thinking that this was a .308 caliber bullet instead of a 7 mm. | ||
<Big50> |
Trajectory at 2 and 300yds is relative to the 100yd zero as Harald pointed out, harmonics are the same unless the barrel has heated and expanded making contact in the barrel channel somewhere, thus throwing impact in ANY number of directions. One other thing, if your 100yds zero was actually anywhere between 70 and 135yds you would be impacting the 100yd target no more than .100" high or low and 200yd impact would be 1.3 - 1.4" low and 300yd impact would be 6.3 - 6.4" low only varying slightly depending on BP at that range. Another note, if you are zeroed at 135yds with bullet in the downward arc it is also zeroed at 70yds in the upward arc. At any rate if you are zeroed at 100yds + or - .100" you are currently without a doubt almost 4" higher at 200yds than you should be, the bullet cannot suddenly at 100yds change directions upward 4MOA to impact this spot 2.5" high. Did the crosshairs break, mount loosen, scope paralax, mirage maybe. I'm sure the problem is quite a simple one when you find it. Sounds like a real shooter though. ------------------ [This message has been edited by Big50 (edited 12-09-2001).] | ||
Powered by Social Strata |
Please Wait. Your request is being processed... |
Visit our on-line store for AR Memorabilia