MOA Reticle Question

Forums General Airgunning MOA Reticle Question

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    josh3rd
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    My reticle is 2 MOA per hash mark. I know MOA @ 100yds is 1.047 or Shooters MOA is 1" IPHY. Ok say chair gun says that at 53yds my rifle which is shooting 18gr jsb's at 748fps the drop will be 3.03" (-5.5moa) which is pretty close. Now the 5.5moa for my reticle will be 2.25moa correct? Because of the 2 MOA per hash mark. I got that part down. But now I entered 15mph into the wind box and it says my drift should be 6". What does that equate to MOA and to my reticle? This is what I think.

    I take 6 inches and multiply it by .53 which equals 3.18 moa and then divide 3.18 by 2 (for 2moa per hash) which equals 1.59. Now is the 1.59 the moa hash I hold on or is my math messed up?

    Now I know chair gun isn't law and I should get dope from actual shooting yada, yada yada I just need to know if I'm doing the math correctly or if I am not and if not, tell, show and or explain the right way please. Thank you

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    fishinwrench
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    Oh God …..Math!  😮

    I'm outta here.

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    LDP
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    Ok say chair gun says that at 53yds my rifle which is shooting 18gr jsb's at 748fps the drop will be 3.03" (-5.5moa) which is pretty close. Now the 5.5moa for my reticle will be 2.25moa correct? Because of the 2 MOA per hash mark.

    It will still be 5.5 moa regardless of your reticle. What I think you are asking is will you hold over 2.25 hash marks not 2.25 moa. If each mark is equal to 2 moa you would hold 2.75 hash marks to equal 5.5 moa. 

     

     

    I take 6 inches and multiply it by .53 which equals 3.18 moa and then divide 3.18 by 2 (for 2moa per hash) which equals 1.59. Now is the 1.59 the moa hash I hold on or is my math messed up?

    I am not sure what you did with your math there but the numbers not correct. If 3" of drop gives 5.5 moa then 6" will be double the moa and you have it as 1.59 moa. So 6" of drift would be more like 11 moa at 53 yards. If it is 11 moa drift you would hold 5.5 hash marks assuming your scopes windage marks are at 2 moa like the verticle marks.

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    Saltlake58
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    I think there's confusion in the way the story problem is written.  and remember I hate math though my wife taught math for many years.

    Maybe one way to look at it is that at 100 yards, 1MOA = 1 Inch.  At 50 Yards, 1 MOA – 1/2 Inch.  So at 100 yards, 1 Hash = 2 inches.  At 50 Yards, 1 hash = 1 inch.

    As I look at the math, I think you multiplied when you should have divided.

    For starters, is the reticle FFP or SFP?  if SFP, what magnification is the standard (most are something like 10X unless otherwise stated, and I assume you are at whatever is standard for your scope)

    Also, for ease of math, for my purposes, 1MOA at 100 yards is 1 inch.  1 MOA at 50 yards is 1/2 inch.

    In this case, each hash on your reticle at 100 yards is 2 inches, at 50 yards is 1 inch.

    If your drop at 50 yards (forget the 53, this is for estimations only) is 3 inches, that's a 6 MOA drop or about a 3 hash drop on your reticle. (3 inch drop  / 1/2 inch per MOA = 6 MOA drop)  Since each hash is 2 MOA, that's a 3 hash drop

    For drift, and I assume the hash marks are also 2 MOA at 100 yards side to side.

    6 inches drift at 50 yards is 12 MOA, or 6 hash marks. At 50 yards 1 inch = 1/2 MOA, so 6 inches drift  / 1/2 inch MOA =12 MOA  If it's 2 MOA per hash, then you have 6 hash drift.

    You are correct, Chairgun is for estimates only.  Shooting is better but Chairgun gets you in the ball park.

    Now, going through this, I've already found errors (corrected before submitting, but still it means I could have errors) if I'm not totally screwed up, take that to the range and test shoot.  Only way to validate is to do the shooting.

    and just remember, I was always lousy at math.

    • This reply was modified 2 months ago by Saltlake58.
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    LDP
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    josh3rd you did your math incorrect for the windage. You multiplied 6 by the yardage. The correct formula for your question about wind drift or drop would be this.

    .53 x 1.047 = .5549 (how many inches equals 1 moa at that range)

    6 / .5549 = 10.81 MOA

    6" = 10.812 moa at 53 yds.

    .53 would be 53 yards and 1.047 is how many inches are in 1 moa at 100 yds. 6 represents the 6" of drift your pellet has at 53 yds.

    As I said earlier in my first post with 10.8 moa drift you would hold 5.5 hash marks if the hash marks were indeed 2 moa. 5.5 would be just a hair more than needed but the actual hold off is 5.4 so it should still be a good hit.

    This is what the formula says your drop is based on a 3.03" drop.

    .53 x 1.047 = .5549

    3.03 / .5549 = 5.460 MOA

    Your hold over for your drop at 53 yds is 5.46 moa so it would be 2.75 hash marks. The actual number would be 2.73 hash marks but its easier to line up exactly between two points and 2.75 hash marks is close enough to the actual 2.73 that the hit should be good.

     

    • This reply was modified 2 months ago by LDP.
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    elh0102
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    The way I figure, if I'm hunting, and my target requires this kind of math to hit, we're both gonna go home safe tonight, and I'll have all my ammo. 

     

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    LDP
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    elh0102

    The way I figure, if I'm hunting, and my target requires this kind of math to hit, we're both gonna go home safe tonight, and I'll have all my ammo. 

     

    You dont need any of the math in field. When you decide you will be venturing into long range shooting whether its airguns or rifles you use a ballistics program or do the math to find out hold over points for drop and wind out to whatever distance you plan on shooting. Once you have the data you go out into the field and shoot at targets to see how accurate your original generated data is. After you shoot in the field and correct anything thats wrong in the theoretical data to the real world data you can make a dope card. You use the dope card while you are in the field shooting. You get better results if you have a log book or multiple dope cards for your rifle and record the differences in the dope you already worked out when the temp changes or altitude. After awhile you will have all the real world data you need to take any shot in any weather and altitude and be confident you will make the hit. 

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    josh3rd
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    LDP

    Ok say chair gun says that at 53yds my rifle which is shooting 18gr jsb's at 748fps the drop will be 3.03" (-5.5moa) which is pretty close. Now the 5.5moa for my reticle will be 2.25moa correct? Because of the 2 MOA per hash mark.

    It will still be 5.5 moa regardless of your reticle. What I think you are asking is will you hold over 2.25 hash marks not 2.25 moa. If each mark is equal to 2 moa you would hold 2.75 hash marks to equal 5.5 moa. 

    LDP, you're right, i miswrote.  It will still be 5.5moa but what I meant to convey is that what will it be on my reticle.  2.75 hash marks as you wrote, correct?

     

    I take 6 inches and multiply it by .53 which equals 3.18 moa and then divide 3.18 by 2 (for 2moa per hash) which equals 1.59. Now is the 1.59 the moa hash I hold on or is my math messed up?

    I am not sure what you did with your math there but the numbers not correct. If 3" of drop gives 5.5 moa then 6" will be double the moa and you have it as 1.59 moa. So 6" of drift would be more like 11 moa at 53 yards. If it is 11 moa drift you would hold 5.5 hash marks assuming your scopes windage marks are at 2 moa like the verticle marks.

    LDP, let me try to clarify this.  Chairgun gave me my drop in inches and moa.  The moa I just divided by 2 and then I knew where my POA on my reticle would be.  Now I should have said this in my original post: What I can't get is that chairgun will give my drift in inches at that given distance.  And this is all on a android phone.  It will say (for example) my drift in inches at 53 yards is 6 inches in a 15mph crosswind.  The speed doesn't matter and I don't want to clutter with other numbers so for the sake of argument its a 6 inch drift at 53yds.  My question is:

    what does that equate to in moa?

    how do i get that answer?

    and where at on my reticle will be the POA?

    Sorry for any and all confusion.  I just need the math

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    josh3rd
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    saltlake, thank you for that.  That seemed so simple and to the others who have replied above his post, my apologies for not totally understanding.

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    josh3rd
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    LDP, you are right as well and for your response to elh's post.  I am shooting a RWS 54 in 22cal shooting jsb 18gr and I'm trying to get my dope cards on point and shooting with the wind blowing 15-20mph is a heck of a tutor

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    josh3rd
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    Thanks again brothers for your help and input. I got it and the formula

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    LDP
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    josh3rd

    Thanks again brothers for your help and input. I got it and the formula

    Glad we could help clarify. I was going to repost my above formula but it looks like you have it and understood it. The formula works good and should get all your starting drops and windage for your dope cards. One piece of advise for adding windage to your dope cards. I usually just put a 10 mph wind column in my dope cards. It keeps the clutter down and its easy to extrapolate the data from there if the wind is different from 10 mph.

    Example:

    Your dope card says a 10 mph wind will be 5 moa at 50 yds.

    Actual wind is 15 mph so it will be 7.5 moa.

    If actual wind was 20 mph then it would be 10 moa.

    That works well for me when figuring out wind drift. That keeps from having multiple rows of wind drift data and causing the dope card to be very large or cluttered. If that doesnt work for you then just change it up and make it usable for you.

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    elh0102
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    I used to do a lot of groundhog hunting, with distances out to 400 yards or so. Here is the problem I see with this whole long distance subject. If your target is at an extended distance, I found it rare that conditions were constant from me to target, which is an assumption for any calculation of drop or wind drift. If you have 3 or 4 different conditions to the target, it is the very rare person who can figure the net effect, assuming they can even be identified. I fully understand the attempt at quantifying the effect of conditions on the shot, I just found that it rarely worked, regardless of perfect math. 

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    LDP
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    elh0102

    I used to do a lot of groundhog hunting, with distances out to 400 yards or so. Here is the problem I see with this whole long distance subject. If your target is at an extended distance, I found it rare that conditions were constant from me to target, which is an assumption for any calculation of drop or wind drift. If you have 3 or 4 different conditions to the target, it is the very rare person who can figure the net effect, assuming they can even be identified. I fully understand the attempt at quantifying the effect of conditions on the shot, I just found that it rarely worked, regardless of perfect math. 

    You are correct that conditions are not constant between you and the target. You are wrong in saying that you cannot use a ballistics program or dope cards to make long range hits consistently. Some people are better at it than others and if you want to be good at it you need to practice and record data in a log book. I would say its really not rare to find people who can make hits at long range. Look at all the people who compete in various long range competitions and you will see its not rare at all. I wouldnt call 400 yds long range so if you had trouble at 400 yds with a 223 or larger cartridge maybe you just need to practice more and study wind dynamics and proper calculating hold over. Allot of people make the mistake of using hold over instead of clicking in their drop in windy conditions. Try holding over 15 moa for drop and also visualizing holding off for wind since the horizontal hash marks will be above the target. Its not easy but if you click in your drop you are able to use your horizontal hash marks for precise wind hold. If it doesnt work for you thats fine but to say it doesnt work is just not true.

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    elh0102
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    You're right, it doesn't work for me. I'm sure it works for those more skilled in its use. As is usually the case, it's operator error. As a gunsmith friend of mine once said, "there's way too much good beer to be drunk to fuss with this". I agree, and I wish the best to all those shooters sufficiently skillful to take advantage of the technology available. 

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    LDP
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    elh0102 also let me clarify that I dont take extremely long range shots at game animals. This isnt a process I would use while elk hunting. I would only shoot beyond 400 yds on game animals if conditions were right and the animal was calm. But I do use dope cards and take long shots on rock chucks, prairie dogs and similar animals. 

    Nothing wrong with doing it your way. We should all make sure we do things the best way we can or what we feel most comfortable with. 

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    josh3rd
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    LDP

    josh3rd

    Thanks again brothers for your help and input. I got it and the formula

    Glad we could help clarify. I was going to repost my above formula but it looks like you have it and understood it. The formula works good and should get all your starting drops and windage for your dope cards. One piece of advise for adding windage to your dope cards. I usually just put a 10 mph wind column in my dope cards. It keeps the clutter down and its easy to extrapolate the data from there if the wind is different from 10 mph.

    Example:

    Your dope card says a 10 mph wind will be 5 moa at 50 yds.

    Actual wind is 15 mph so it will be 7.5 moa.

    If actual wind was 20 mph then it would be 10 moa.

    That works well for me when figuring out wind drift. That keeps from having multiple rows of wind drift data and causing the dope card to be very large or cluttered. If that doesnt work for you then just change it up and make it usable for you.

    LDP Damn good idea…a happy medium

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    josh3rd
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    elh0102

    I used to do a lot of groundhog hunting, with distances out to 400 yards or so. Here is the problem I see with this whole long distance subject. If your target is at an extended distance, I found it rare that conditions were constant from me to target, which is an assumption for any calculation of drop or wind drift. If you have 3 or 4 different conditions to the target, it is the very rare person who can figure the net effect, assuming they can even be identified. I fully understand the attempt at quantifying the effect of conditions on the shot, I just found that it rarely worked, regardless of perfect math. 

    Understood elh. Thank you

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