THE FOLLOWING POST IS INCORRECT. I HAVE LEFT IT HERE JUST FOR REFERENCE TO THE LATER POST. CHEERS
Good day,
Being new to the long distance precision shooting world there are a million questions that run though my mind every time I think about this sport. I am sure there are more people out there thinking the exact same thing so maybe we can clarify some things.
I have been shooting at 100 yards (100 Yard Zero) for some time but I am starting to shoot further now. I think I understand the concept of Bullet drop vs. Clicks but I was hoping I could run it by you guys to see if my logic adds up.
First off I am running a Bushnell 3200 5x15 with target knobs on a Tikka T3. The scope is set at ¼ MOA increments.
So the way I see it, 1 MOA @ 100 yards equals 4 clicks. The target knobs have 12 MOA per rotation which works out to 48 clicks at 100 yards. The scope is said to have 50 MOA all together which should mean that I can do a total of 4 full rotations and 2MOA and no clicks (which I will refer to as 4R 2M 0C). Hopefully I am doing fine with my math so far.
So I am going to assume we are shooting a .308 match bullet (I am just going to make up drop values to make this easier) and that the bullet drop goes as follows:
Zero’d @ 100 yards
-0.5” @ 200 yards
-1.5”@ 300 yards
-3.5 @400 yards
-5.0” @ 500 yards
Blah Blah Blah
-25”@1000 yards
I realize these are probably really unrealistic but again it’s just to make the example easier to work with.
So let’s say I am sitting at the shooters bench now zero’d at 100 yards and I want to shoot at a target 300 yards away. I look up my bullet drop and notice that it drops -1.5”. So if it takes 4 clicks to move a bullet 1” @ 100 yards, now it will take 12 clicks (4clicks X 3 = 12) to move that same bullet over 1” at 300 yards. Since we are shooting at 300y and the drop is 1.5” we calculate to 18 clicks OR 4M 2C.
So let’s try a 500 yard shot. It would take 20 clicks to shift a bullet 1” at 500 yards. We notice that bullet drop for 500y is -5.0” so 20x5=100 clicks OR 2R 1M 0C.
Ok one last one.
Its time for the last shot. 1 Bullet left! But the target is 1000 yards away. So 40 clicks shifts a bullet 1” at 1000 yards. So 40x25=1000 clicks
48 clicks goes into 1000 clicks 20 times and your left with 40 clicks divided by 4 (amount of clicks) = 10. So you would have to Adjust 1000 clicks OR 20R 10M
Seeing as my scope only has 4R 2M 0C worth of adjustment I would not be able to make that shot without using a shim or re-zeroing at a longer distance to reduce the amount of bullet drop.
Does that make sense or is my math way off?
Cheers
Good day,
Being new to the long distance precision shooting world there are a million questions that run though my mind every time I think about this sport. I am sure there are more people out there thinking the exact same thing so maybe we can clarify some things.
I have been shooting at 100 yards (100 Yard Zero) for some time but I am starting to shoot further now. I think I understand the concept of Bullet drop vs. Clicks but I was hoping I could run it by you guys to see if my logic adds up.
First off I am running a Bushnell 3200 5x15 with target knobs on a Tikka T3. The scope is set at ¼ MOA increments.
So the way I see it, 1 MOA @ 100 yards equals 4 clicks. The target knobs have 12 MOA per rotation which works out to 48 clicks at 100 yards. The scope is said to have 50 MOA all together which should mean that I can do a total of 4 full rotations and 2MOA and no clicks (which I will refer to as 4R 2M 0C). Hopefully I am doing fine with my math so far.
So I am going to assume we are shooting a .308 match bullet (I am just going to make up drop values to make this easier) and that the bullet drop goes as follows:
Zero’d @ 100 yards
-0.5” @ 200 yards
-1.5”@ 300 yards
-3.5 @400 yards
-5.0” @ 500 yards
Blah Blah Blah
-25”@1000 yards
I realize these are probably really unrealistic but again it’s just to make the example easier to work with.
So let’s say I am sitting at the shooters bench now zero’d at 100 yards and I want to shoot at a target 300 yards away. I look up my bullet drop and notice that it drops -1.5”. So if it takes 4 clicks to move a bullet 1” @ 100 yards, now it will take 12 clicks (4clicks X 3 = 12) to move that same bullet over 1” at 300 yards. Since we are shooting at 300y and the drop is 1.5” we calculate to 18 clicks OR 4M 2C.
So let’s try a 500 yard shot. It would take 20 clicks to shift a bullet 1” at 500 yards. We notice that bullet drop for 500y is -5.0” so 20x5=100 clicks OR 2R 1M 0C.
Ok one last one.
Its time for the last shot. 1 Bullet left! But the target is 1000 yards away. So 40 clicks shifts a bullet 1” at 1000 yards. So 40x25=1000 clicks
48 clicks goes into 1000 clicks 20 times and your left with 40 clicks divided by 4 (amount of clicks) = 10. So you would have to Adjust 1000 clicks OR 20R 10M
Seeing as my scope only has 4R 2M 0C worth of adjustment I would not be able to make that shot without using a shim or re-zeroing at a longer distance to reduce the amount of bullet drop.
Does that make sense or is my math way off?
Cheers
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