This dude is UNBELIEVABLE! i cant seem to find the ten thumbs up icon? in my twenty plus years of shooting long range (and winging it) he has clarified all the mystery in about an hour. What an incredible instructor!
Sean Allyn 8 months after your comment and I still can't find any better than this guy on TH-cam. Brilliant, explains everything do well and simple to follow
I just started watching this channel today and already learned alot..i like his cut through the bs and tries to explain things in layman’s terms so people can understand it
5:59 umm no I had a much better math teacher than this... the angles of a triangle add to 180• A right angle is 90• so the other two angles are equal to 90• supplementary.... therefore the two angles can’t be equal unless they are both 45• It doesn’t fuck up his theorem. But he was incorrect on that aspect.
@@emiliochavez5121 He didn't say the 2 angles INSIDE the triangles were equal. He said the angle between the hypotenuse line and the imaginary line perpendicular to the opposite side is the same as the angle between the hypotenuse line and the adjacent line. Which is correct and there are better ways to prove it but the simplest is that the angles inside the triangle (in this case) are [40 degrees + 90 degrees + (90 degrees - 40 degrees)], which add up to 180 degrees. Now I apologize for my improper terminologies, cause I forgot all of them.
@JKay11235 a^2 + b^2 = c^2 If c = 20 ft (c being the ladder) and b = 3 ft (the base) then it goes: a^2 + 3^2 = 20^2 Then you subtract 3^2 from 20^2 and then square root to find a (which is 19.77 feet)
@JKay11235 Math and Science *can* be corrupted by a political agenda, have you heard of p-hacking? Also math is not facts and laws about the universe, it's merely a conceptual framework that may not even have to be related to reality at all, for example 4-D+ spacial dimensions. Science is *not* "facts and laws of the universe" although it makes an attempt at being so it can only be a _close approximation to it_ .
I learned sines and cosines in HS about 35 years ago and even after all these years listening to this guy makes it all clearer than ever!! If only my teachers would have put it in shooting terms and real world situations it would have made so much more sense and made it interesting for everyone especially the guys!! They would have been eager to learn it then!! He really should make a video series for schools to use!!
I just got out of the Army Infantry. If I were to stay in I would have showed my soldiers your videos. This is some very helpful information for anyone shooting at distances not just snipers. The Army never taught me these details of shooting. I had to research and figure it out on my own. The Army would greatly benefit it they signed this dude up for some educational videos. Not that their leaders and drill sergeants aren't good, its just that many are not taught themselves how to teach and train soldiers. This guy does a fantastic job of teaching.
Bullshit! Your soldiers’ would not have understood most of what he said. He is teaching way above a typical person’s, let alone an infantryman’s, level of comprehension.
@@lextalionis3754 lol I mean I understood what he was talking about. It isn't rocket science, its typical high school level math. The point though isnt putting an infantryman through a math class. It's giving them an idea what their rounds are doing outside of a flat range and why.
So glad I found this channel. I've been shooting for 45 years and this guy is the best I've ever seen explaining these concepts. This is a long way from years ago where you'd get to know your rifle and scope and guestimate hold over and angles.
Ryan's videos are the best. He explains everything in such a logical and easy way to understand. I am new at long range and am so happy that I found these videos so I can start off the right way and not suffer too much while I am learning.
An inclinometer on your compass makes for a really quick angle reference. Remembering the cosine formula makes this pretty easy. Carrying a solar calculator as well can help a lot or at least a cheat sheet.
Ryan, you really are a great instructor. I hope you write a book some day. I'd be happy to be a reviewer ;-) And, once again, thank you for your service to our country.
Great explanation, feel like I just sat through a class. 2 things I’m not 100% clear on: 1. With the ranging using the mil reticle in the scope issue, you are saying that one would use the same cosign used in the bullet drop angle issue as to target size? And then just use/calculate it again for the bullet drop? 2. Good high-end laser rangefinders will have a setting that adjusts the yardage to true yardage based on the slope angle. If I have one of these and it gives me the reduced number, can I just treat that as accurate for shooting purposes? Problem solved?
Very thorough, and easy to understand explanation. I went up to a new spot today, to start coyote hunting. It's on the top of a huge hill, looking down into a field(lots of rabbits). I have an old range finder that shows its about 125 yards to the bottom, but it's a pretty steep angle down, I need to figure out what that angle is, and compensate for it. This video will help me a lot. Thanks! I don't have any fancy tools, but I think I can use a level, and a carpenters square, and figure it out pretty close. I'll sit my rifle on a table, point it at the bottom of the hill, place the level on my barrel, then use the square to figure out the angle. I'm thinking it's about 30-35 degrees.
The fact that there are so many vets in the comments saying "you explain this better than the military ever did" makes me think you should be selling these videos lol
No need for angle measuring tools. Just use the width to get the range to the target because it doesn’t change on an incline. Then get the height of the target from that perspective and divide by the known height to get the cosine of the angle. E.g. you use the width and determine the target is 100 yards away, you then see the height of the target is 8.8 moa. At 100 yards that’s 9.2”, divide by the known target height of 12” gives you ~0.77, or arccos(40°). Therefore, the target is on a 40° incline.
my rangefinder (sig, least expensive one) has an option to adjust range for angle automatically. yay. that said, excellent explanation, perfectly well paced.
Ryan, you're a great teacher and great speaker. If my geometry teacher had taught us about bullet drop and milradians, I might have learned a lot more math.
Great vid . I have for decades used a surveyors pocket transit for finding angles . I also use the width and not height as You do for the exact reason . Having those practical cosigns make so much sense . Thank You Sir
This perfectly explains why I had a higher hit on a deer than I expected. My range estimation was only a touch short, but I didn't consider the angle at all and hit almost 6 inches higher than I should have if my range was correct.
Ryan, as with all of your videos great stuff, very informative. In the video you speak about using COS when shooting down a slope and how shooting directly down i.e. off a cliff you do not compensate in any way. What about shooting up a slope, do you use the same figure?
Very awesome video great to use especially if your range finder is broken or doesn't account for angles. A range finder that has an archery mode it dose all this for you. But still 10 thumbs up great video!
The better way to explain it is that the gravity is a vector which has a component of the force pulling the bullet along the flight path. This means it’s velocity is higher and bullet drop less.
It’s interesting how we instinctively recognize these rules in some situations, but not others. Anyone who ever threw a baseball or football to someone else who was waaaay down a huge hill knew instinctively that they only had to lob the ball without much effort to get it to the guy at the bottom.
Hi, you said - Distance X cosine = Angle range and - Mil Range X cosine = Angle range Therefor - Distance X cosine = Mil Range X cosine Or - Distance = Mil Range So you can use Mil range for Distance directly - no problem. Note this is for vertical Mil range only not horizontal Mil ranges.
Terrific video, i learned alot.. It occurred to me that the "distance" i get from my targets perceived height in the reticle divided into the "true" distance achieved from the width gives me the cosine; then the "true distance" times this value is the distance needed to adjust for drop, just a thought. Maybe american youth would learn trig better if they could use shooting examples.
1 equation: milrange * cosine value^2 = angle range. If the animal is not standing perpendicularly to you, this would introduce another need to perform a cosign calculation. This poses an interesting scenario where the back haunches to the front shoulders are assumed to be flat.
this is good between Flat and 45 deg off Flat either up or down. But once you get past 45 deg you have other factors that come into play. Such as the fact that your Scope is not directly in line with your barrel It is infact aprox 2" above your barrel and therefore always angled a bit ""Down"" naturally. He mentioned that in nature you will not find many Angles greater than 30deg and this is true if your on a hill or mountain (without cliff) and aiming "Down" yet this is not the normal scenario we all experience this. Infact it is very normal for people to be aiming at small game or birds that are perched in a tree that you are various distance from. Therefore the angle can be much more than 45deg in alot of scenarios. This is where you need to know your first zero as well as 2nd zero and centerline diff between your scope and barrel. Gets very tricky after 45 deg, and this simple method is not nearly enough.
I'm not totally understanding your question, however, the cosine numbers have nothing to do with the yards on the white board. The cosine number deals with the angle, not the distance. That being said, the cosine numbers on the white board will stay the same with the ANGLE written down on the white board (10=.98, 20=.94, 30=.87 etc), because that cosine is based upon that particular angle. I hope this helps with your question.
I was looking at a ballistic calculator of my chosen round and found that something is off. In the example if you had a total of 400 yards to the target and 308 yards is your "x-component" (tangent to the surface of the earth from you and the shooter). I'm having trouble with the conceptual understanding of it. The time of flight would be the same as if you are shooting level with the target at the same distance. so gravity is going to act on it for the same amount of time as if you were shooting level (in my case .53 seconds). Remember that gravity is 32 feet per second per second so in the level calculation my bullet will drop 54 inches (from true level, from the exit of the muzzle, assuming I fired it straight and not at an angle like all rifles do (this isn't from my zero, from my zero it's 24 inches)). Now I know for a fact that the bullet drop isn't the same in your example as if you were shooting at a target at 308 yards, because if you were shooting a target at 308 yards the time that it will be affected by gravity would be shorter than that angled shot. I think how reality works is that the bullet (being shot at an angle) now has gravity affected by an angle. This means we have to separate them into a "x and y" component (components that your scope can measure) so we can be able to hit the target. The factors causing the bullet to go down is the "y" component of the velocity of the bullet since you're aiming it down, and the "y" component of gravity. The "x" components causing your bullet to go towards the target is the full velocity of the bullet (since you're pointing it at the target, _mostly_) and the "x" component of gravity _accelerating towards your target_ (strangely the balistic calculator says it's the same flight time but _I think_ it's slightly shorter because of this, or maybe not air friction might cancel this factor just right in this case). I might've messed up with the reference frame a couple of times and I'm taking about two different 2-d planes but that's my general idea of how shooting a target at a 40 degree angle decline at 400 yards is completely different than shooting a target at 308 yards level. That said your process might very well work but conceptually it seems off (to me). The angle of the bullet to the center of gravity of earth is 40 degrees so the cosine of 40 by the normally 54 inch drop is about 41.6 inches. Now, it obviously didn't hit higher, you're shooting at an angle of decline so the path of the bullet is a lot more lower, the elevation from you to the target to be exact, but in relation to the "true level, from the exit of the muzzle" it will drop 41.6 inches. The 308 yard level shot may also have dropped about 41.6 inches but due to the different nature of shooting at a decline the bullet trajectories are completely different due to the the factors that I broke into components listed above. I would like to have feedback on your understanding of this physics and whether or not these actually are worth considering (or they are negligible). I believe that conceptually we should look at it as the forces acting on the bullet from the distance exiting the muzzle (at ~2700 feet per second) to impacting your target. practically it would be easier just to understand cosine angles and multiplying that by the distance from your target measured by the moa's in your scope and assumed 5ft height of the target. Since you already know your dope you're good to go. I saw you use the calculator for the cosines so I assume you don't know it in the field. What do you do if you do shoot out in the field(mountains) without a calculator?
The width of the target doesn't change with your up/down perspective but does change if the target isn't broadside to you. If the, lets say, deer was 45 degrees off of broadside, the apparent width will be much smaller. I imagine seeing that through a scope at 500yds would be quite difficult but maybe I just need to buy a better scope... :)
John Kidd Your right on with this point. The reason height is usually used in the first place is because width is rarely presented as 'squarely' as height tends to be. Using width hardly "solves all your problems."Good to have another option I guess. Honestly theres a fair amount of 'eyeballing' and 'guess-work' with this whole method of estimating range. If the animal your measuring is 20% larger or smaller than the average you are calculating from, its going to throw your range estimate off by 20%. At 500m you might still hit, at 800m thats enough to miss entirely.
There is a thing called included angle where you measure both height and width and add them together and plug into this formula for ranging; D(KM)=H(M)/K, where K is the included angle. I think this method can comensate for the apparent foreshortening of an angled target
I hope this video will help in the next hunting season, i missed shooting an elk at 301yards at -29° I click 3.2MOA.... and missed 5 shot at the elk. I did this math is should be 263yards, and that's .7 MOA.🤔
Better than any teacher i've ever had in school LOL Copy and paste cheat sheet if anyone wants it. 5°angle = 0.99 or 99% 10°angle = 0.98 or 98% 20°angle = 0.94 or 94% 30°angle = 0.87 or 87% 40°angle = 0.77 or 77% 50°angle = 0.64 or 64% 60°angle = 0.50 or 50% 70°angle = 0.34 or 34% 80°angle = 0.17 or 17% 90°angle = 0 or 0%
This had been one of the best series I have ever watched. Would have loved to had you as teacher. Thanks so much for such a thorough and simple presentation on a complex and dynamic concept ( shooting ).
If you're shooting downhill can you just look at your barrel and guesstimate what angle it's tilted at in comparison to a straight imaginary line? Great video!
Did you guys notice how a 60 degree angle (not 45 degrees) gives you a .5 Cosine.. I always though a 45 degree would give you half the distance. Glad to learn about these Cosines.
So Ryan, in shooting at an angle, with target downhill (in this example) are you saying you simply substitute your mil height with mil width in the "Mil Estimation" formula, right from the start and then what step would that allow you to skip, the "Angle Range" formula. I've tuned into your videos time and time again, you have an impressive ability in delivering concise and clear subject matter. Great videos.
Very nice video. One thing tho... What about if I'm shooting uphill ? I understand from your logic that the bullet is being pull to earth harder so the true range is longer than the actual range. The question is : where can I find a chart with cosine for both uphill and downhill angles? I guess the cosine for uphill will start from 1.01 and beyond. I do have a rangefinder that has a built in inclineometer that shows the angle in degrees. Thank you for the videos and taking the time to respond.
If my teachers in HS would have explained tan cos and sin the way he has I would have understood it completely!!! They just tell you the what without showing you the why and drawing a picture and explaining it in real life terms. This guy should make math videos for schools!!! He could keep so many kids from just being lost!!
It would seem that circles (tires) will always be seen on angle as an ellipse, with the greatest length indicating its diameter. Then it remains to the shooter to decide what vehicles have what diameter tires.
a2 + b2 = c2. you are shooting down the hypotenuse of a triangle. if you know your height above a target ( a2 ) and the distance (c2) to the target you can figure out what the bullet thinks the distance is to the target (b2).
Loving your videos bro! you should strike out on your own?!?!? I just finished watching you "understanding MOA" and it was an eye opener for me as i have "never" fired a gun that had a scope on it and just recently purchased a Centerpoint 4x16x44 and still have to mount it.Which i will go back and watch your video on that as well!? :>) Nothings better then a fellow Vet showing people how to shoot.I was in the Navy and only had to qualify with a .22Cal 1911 as well as a .22Cal rifle.As a teen i use to hunt/trap shoot and loved it but after school when i entered the service my life got very busy.I never picked up a firearm again until recently.bout 7yrs ago i decided after a few home invasions in my area that we needed a firearm in the house for our own protection.It didnt stop there though i purchase a bolt gun soon after .I purchased a Savage Axis in 6.5Creedmoor and it has very quickly become my favorite cartridge!:>)Although soon after i have started collecting ARs and their variants! I am up to 5 in total.Two M4s which are my fav. Model!!! But i also have a .224Valkrie and one in 7.62x39 and another in 6.5Grendel.If you cant tell by my purchases i am very much into long range shooting!Im looking to as soon as possible to make a shot at 1mile!!!!!! But in my bluest state of N.J. there wasnt much hope for that.:>/ but just two days we had legislation past that makes my county in N.J. a 2A sanctuary!!!:>) so my hopes have risen!!!:>)I still have to travel out of state to shoot further then 300 or 400yrds!:>/ I thanks you for your content and info i'm grateful! Also i want to thank you for your service!
I attended Army Sniper School in 2004, And this is the best explanation I have heard, and actually understood. Thank you.
Same for me
2009. I agree.
I kept falling asleep in class. (1988) It's like he was talking Greek to me. LOL
@@toddjenest3212 I was an instructor at the sniper school in 1988. We may have run into each other.
He did explain really well but as soon as he mentioned trig and started drawing formulas..my brain shut off..i could never understand that shit
This dude is UNBELIEVABLE! i cant seem to find the ten thumbs up icon? in my twenty plus years of shooting long range (and winging it) he has clarified all the mystery in about an hour. What an incredible instructor!
Sean Allyn 8 months after your comment and I still can't find any better than this guy on TH-cam. Brilliant, explains everything do well and simple to follow
Sean Allyn 1 thumbs up will do ya! 🤣
I just started watching this channel today and already learned alot..i like his cut through the bs and tries to explain things in layman’s terms so people can understand it
Yes, this is a very good presentation.
@@NightStalkerDNS😅zadaq😢😅t😊s
Better math teacher than i had in school
😂😂😂
5:59 umm no I had a much better math teacher than this... the angles of a triangle add to 180•
A right angle is 90• so the other two angles are equal to 90• supplementary.... therefore the two angles can’t be equal unless they are both 45•
It doesn’t fuck up his theorem. But he was incorrect on that aspect.
@@emiliochavez5121 no he was not.. He pointed to the adjacent corner
@@emiliochavez5121 He didn't say the 2 angles INSIDE the triangles were equal. He said the angle between the hypotenuse line and the imaginary line perpendicular to the opposite side is the same as the angle between the hypotenuse line and the adjacent line. Which is correct and there are better ways to prove it but the simplest is that the angles inside the triangle (in this case) are [40 degrees + 90 degrees + (90 degrees - 40 degrees)], which add up to 180 degrees. Now I apologize for my improper terminologies, cause I forgot all of them.
Dont matter to me ive never understood this shit
I learned more relevant math in this video than all my math classes in high school combined.
+AJ Day Then, it's evident that either: You didn't listen, or, your schooling system failed you. lol
@JKay11235 That's 19.77 feet (still generally close to 20)
@JKay11235 a^2 + b^2 = c^2
If c = 20 ft (c being the ladder) and b = 3 ft (the base) then it goes: a^2 + 3^2 = 20^2
Then you subtract 3^2 from 20^2 and then square root to find a (which is 19.77 feet)
@JKay11235 Math and Science *can* be corrupted by a political agenda, have you heard of p-hacking? Also math is not facts and laws about the universe, it's merely a conceptual framework that may not even have to be related to reality at all, for example 4-D+ spacial dimensions. Science is *not* "facts and laws of the universe" although it makes an attempt at being so it can only be a _close approximation to it_ .
Your class brought back to me those classes I sat in as a Marine LCpl over 50 years ago. Thank you. Numbers don’t change.
I learned sines and cosines in HS about 35 years ago and even after all these years listening to this guy makes it all clearer than ever!! If only my teachers would have put it in shooting terms and real world situations it would have made so much more sense and made it interesting for everyone especially the guys!! They would have been eager to learn it then!! He really should make a video series for schools to use!!
I just got out of the Army Infantry. If I were to stay in I would have showed my soldiers your videos. This is some very helpful information for anyone shooting at distances not just snipers. The Army never taught me these details of shooting. I had to research and figure it out on my own. The Army would greatly benefit it they signed this dude up for some educational videos. Not that their leaders and drill sergeants aren't good, its just that many are not taught themselves how to teach and train soldiers. This guy does a fantastic job of teaching.
He is a former US Army Scout/Sniper
Bullshit! Your soldiers’ would not have understood most of what he said. He is teaching way above a typical person’s, let alone an infantryman’s, level of comprehension.
@@lextalionis3754 lol I mean I understood what he was talking about. It isn't rocket science, its typical high school level math. The point though isnt putting an infantryman through a math class. It's giving them an idea what their rounds are doing outside of a flat range and why.
Excellent teaching technique, clear ,concise, effective.
also best explanation of mils/angles/wind etc I've seen yet.
By far this was the best vid I saw ...short n sweet ...crisp n brief ... awesome ...thank you mate
Thanks a million for explaining this perfectly 👍🎯
Now I have a true appreciation of Trigonometry. Great instruction.
So glad I found this channel. I've been shooting for 45 years and this guy is the best I've ever seen explaining these concepts. This is a long way from years ago where you'd get to know your rifle and scope and guestimate hold over and angles.
Ryan's videos are the best. He explains everything in such a logical and easy way to understand. I am new at long range and am so happy that I found these videos so I can start off the right way and not suffer too much while I am learning.
An inclinometer on your compass makes for a really quick angle reference. Remembering the cosine formula makes this pretty easy. Carrying a solar calculator as well can help a lot or at least a cheat sheet.
thank you sire your videos have been so helpful in my education!
Just a fantastic video! You gained a subscriber and I’m sharing this with my my long range crew
Thank you for the amazing feedback!
Ryan, you really are a great instructor. I hope you write a book some day. I'd be happy to be a reviewer ;-)
And, once again, thank you for your service to our country.
He's written a few👍
Outstanding! I couldn't imagine a way that this information could be put into a video any more effectively. Thank you very much!
Outstanding job Ryan! Whatever you’re being paid… you deserve a raise!
Great explanation, feel like I just sat through a class. 2 things I’m not 100% clear on:
1. With the ranging using the mil reticle in the scope issue, you are saying that one would use the same cosign used in the bullet drop angle issue as to target size? And then just use/calculate it again for the bullet drop?
2. Good high-end laser rangefinders will have a setting that adjusts the yardage to true yardage based on the slope angle. If I have one of these and it gives me the reduced number, can I just treat that as accurate for shooting purposes? Problem solved?
Very thorough, and easy to understand explanation. I went up to a new spot today, to start coyote hunting. It's on the top of a huge hill, looking down into a field(lots of rabbits). I have an old range finder that shows its about 125 yards to the bottom, but it's a pretty steep angle down, I need to figure out what that angle is, and compensate for it. This video will help me a lot. Thanks! I don't have any fancy tools, but I think I can use a level, and a carpenters square, and figure it out pretty close. I'll sit my rifle on a table, point it at the bottom of the hill, place the level on my barrel, then use the square to figure out the angle. I'm thinking it's about 30-35 degrees.
This guys one heck of an instructor. Well done sir
Great instructional video on a subject that is not very common
hands down, best video I've seen on the topic
The fact that there are so many vets in the comments saying "you explain this better than the military ever did" makes me think you should be selling these videos lol
3 yrs of college, i coulda watch this guy and saved all that money
Nothing that others haven't sayed already, but. This guy is amazing.
Talk about the KISS principle...thanks for sharing this!
"I'll never use this trigonometry stuff again in my life!" , said everyone.
How about now?
lol I'd prefer this 17 minute video to a year in math class
Triggernometry
No need for angle measuring tools. Just use the width to get the range to the target because it doesn’t change on an incline. Then get the height of the target from that perspective and divide by the known height to get the cosine of the angle. E.g. you use the width and determine the target is 100 yards away, you then see the height of the target is 8.8 moa. At 100 yards that’s 9.2”, divide by the known target height of 12” gives you ~0.77, or arccos(40°). Therefore, the target is on a 40° incline.
my rangefinder (sig, least expensive one) has an option to adjust range for angle automatically. yay.
that said, excellent explanation, perfectly well paced.
These are premium videos for free on TH-cam. Please keep these up.
Thanks.
Best explanation to the angle shooting mystery ever..Thanks.
Ryan, you're a great teacher and great speaker. If my geometry teacher had taught us about bullet drop and milradians, I might have learned a lot more math.
Great vid . I have for decades used a surveyors pocket transit for finding angles . I also use the width and not height as You do for the exact reason . Having those practical cosigns make so much sense . Thank You Sir
I love this guy ... Best instructor ever 💙
I have learned more from you than any other source ever. You should think about doing this professionally!
Fantastic Ryan, and Thanks. You truly have a gift for teaching. I'll bet you really love what you do brother.
great videos wish i could have him as a personal tutor in long range shooting
If I didn't already know trig pretty well, I'd want to learn it from Ryan.
This perfectly explains why I had a higher hit on a deer than I expected. My range estimation was only a touch short, but I didn't consider the angle at all and hit almost 6 inches higher than I should have if my range was correct.
Exactly what i was missing since my Sniper Course. You're an awesome teacher. Now i have to practice !
Ryan, as with all of your videos great stuff, very informative. In the video you speak about using COS when shooting down a slope and how shooting directly down i.e. off a cliff you do not compensate in any way. What about shooting up a slope, do you use the same figure?
Ryan, Sir, thank´s a lot for this free videos on youtube, fantastic tips!
Very awesome video great to use especially if your range finder is broken or doesn't account for angles. A range finder that has an archery mode it dose all this for you. But still 10 thumbs up great video!
Another bit of the puzzle solved. Thanks
12 years old and still a great video.
Great teacher. You rock.
finally a clear explanation, thanks again
Ryan, very informative video, like all your others. Clear and easy to understand. thanks
Finally...I now know what Pithagoras was thinking about...Loove the guy!
What a great and informative tutorial. Thank you,
The better way to explain it is that the gravity is a vector which has a component of the force pulling the bullet along the flight path. This means it’s velocity is higher and bullet drop less.
Awesome! I hunt on hills sometimes so this could be very useful. I'm planning on getting a good grasp on angles soon. Thank you very much!
Thank you for the video, it's very useful & helpful. This was the last formula that I needed for calculating a long range shot that I didn't have.
Really nice explanation.
Ease of instruction
Thank you.
Great instruction!
Absolutely excellent explanation, Thx Ryan!
It’s interesting how we instinctively recognize these rules in some situations, but not others. Anyone who ever threw a baseball or football to someone else who was waaaay down a huge hill knew instinctively that they only had to lob the ball without much effort to get it to the guy at the bottom.
This is AMAZING video. Thanks a lot.
Hi,
you said - Distance X cosine = Angle range
and - Mil Range X cosine = Angle range
Therefor - Distance X cosine = Mil Range X cosine
Or - Distance = Mil Range
So you can use Mil range for Distance directly - no problem. Note this is for vertical Mil range only not horizontal Mil ranges.
I've been in the marine corps for 9 years and this is by far the best explanation I've ever heard regarding range vs elevation
thank you for your service Bo
Your vids are fantastic, very clear and easy to understand. Thank you very much for making me better !
SO cool! I love how physics and math is incorporated in all this.
This guys great. I haven't seen a new vid from him in a while though..
Great video explained in layman's terms.
Terrific video, i learned alot.. It occurred to me that the "distance" i get from my targets perceived height in the reticle divided into the "true" distance achieved from the width gives me the cosine; then the "true distance" times this value is the distance needed to adjust for drop, just a thought. Maybe american youth would learn trig better if they could use shooting examples.
1 equation: milrange * cosine value^2 = angle range. If the animal is not standing perpendicularly to you, this would introduce another need to perform a cosign calculation.
This poses an interesting scenario where the back haunches to the front shoulders are assumed to be flat.
this is good between Flat and 45 deg off Flat either up or down. But once you get past 45 deg you have other factors that come into play. Such as the fact that your Scope is not directly in line with your barrel It is infact aprox 2" above your barrel and therefore always angled a bit ""Down"" naturally. He mentioned that in nature you will not find many Angles greater than 30deg and this is true if your on a hill or mountain (without cliff) and aiming "Down" yet this is not the normal scenario we all experience this. Infact it is very normal for people to be aiming at small game or birds that are perched in a tree that you are various distance from. Therefore the angle can be much more than 45deg in alot of scenarios. This is where you need to know your first zero as well as 2nd zero and centerline diff between your scope and barrel. Gets very tricky after 45 deg, and this simple method is not nearly enough.
At 7:45 in the video
Do the cosine numbers always stay the same with the yards on the white board???
Great video
Thank you
I'm not totally understanding your question, however, the cosine numbers have nothing to do with the yards on the white board.
The cosine number deals with the angle, not the distance.
That being said, the cosine numbers on the white board will stay the same with the ANGLE written down on the white board (10=.98, 20=.94, 30=.87 etc), because that cosine is based upon that particular angle.
I hope this helps with your question.
You answered my question. Thank you!
great explanation 👍
You are an amazing instructor !!
wow, these videos help me a lot. thanks man, keep it up.
I was looking at a ballistic calculator of my chosen round and found that something is off. In the example if you had a total of 400 yards to the target and 308 yards is your "x-component" (tangent to the surface of the earth from you and the shooter).
I'm having trouble with the conceptual understanding of it. The time of flight would be the same as if you are shooting level with the target at the same distance. so gravity is going to act on it for the same amount of time as if you were shooting level (in my case .53 seconds). Remember that gravity is 32 feet per second per second so in the level calculation my bullet will drop 54 inches (from true level, from the exit of the muzzle, assuming I fired it straight and not at an angle like all rifles do (this isn't from my zero, from my zero it's 24 inches)). Now I know for a fact that the bullet drop isn't the same in your example as if you were shooting at a target at 308 yards, because if you were shooting a target at 308 yards the time that it will be affected by gravity would be shorter than that angled shot.
I think how reality works is that the bullet (being shot at an angle) now has gravity affected by an angle. This means we have to separate them into a "x and y" component (components that your scope can measure) so we can be able to hit the target. The factors causing the bullet to go down is the "y" component of the velocity of the bullet since you're aiming it down, and the "y" component of gravity. The "x" components causing your bullet to go towards the target is the full velocity of the bullet (since you're pointing it at the target, _mostly_) and the "x" component of gravity _accelerating towards your target_ (strangely the balistic calculator says it's the same flight time but _I think_ it's slightly shorter because of this, or maybe not air friction might cancel this factor just right in this case). I might've messed up with the reference frame a couple of times and I'm taking about two different 2-d planes but that's my general idea of how shooting a target at a 40 degree angle decline at 400 yards is completely different than shooting a target at 308 yards level. That said your process might very well work but conceptually it seems off (to me). The angle of the bullet to the center of gravity of earth is 40 degrees so the cosine of 40 by the normally 54 inch drop is about 41.6 inches. Now, it obviously didn't hit higher, you're shooting at an angle of decline so the path of the bullet is a lot more lower, the elevation from you to the target to be exact, but in relation to the "true level, from the exit of the muzzle" it will drop 41.6 inches. The 308 yard level shot may also have dropped about 41.6 inches but due to the different nature of shooting at a decline the bullet trajectories are completely different due to the the factors that I broke into components listed above.
I would like to have feedback on your understanding of this physics and whether or not these actually are worth considering (or they are negligible). I believe that conceptually we should look at it as the forces acting on the bullet from the distance exiting the muzzle (at ~2700 feet per second) to impacting your target. practically it would be easier just to understand cosine angles and multiplying that by the distance from your target measured by the moa's in your scope and assumed 5ft height of the target. Since you already know your dope you're good to go. I saw you use the calculator for the cosines so I assume you don't know it in the field. What do you do if you do shoot out in the field(mountains) without a calculator?
eidt: you explain what you used in the field. that's pretty useful. an angle cosine indicator?
paragraph breaks? jesus...
The width of the target doesn't change with your up/down perspective but does change if the target isn't broadside to you. If the, lets say, deer was 45 degrees off of broadside, the apparent width will be much smaller. I imagine seeing that through a scope at 500yds would be quite difficult but maybe I just need to buy a better scope... :)
John Kidd Your right on with this point. The reason height is usually used in the first place is because width is rarely presented as 'squarely' as height tends to be. Using width hardly "solves all your problems."Good to have another option I guess. Honestly theres a fair amount of 'eyeballing' and 'guess-work' with this whole method of estimating range. If the animal your measuring is 20% larger or smaller than the average you are calculating from, its going to throw your range estimate off by 20%. At 500m you might still hit, at 800m thats enough to miss entirely.
John Kidd If you just remember that you'll miss high when shooting at angles, you'll probably be just fine hunting.
There is a thing called included angle where you measure both height and width and add them together and plug into this formula for ranging;
D(KM)=H(M)/K, where K is the included angle. I think this method can comensate for the apparent foreshortening of an angled target
Great video.about a video on Bullet drop compensation on Nikon prostate 5 .thank you Ryan.
I hope this video will help in the next hunting season, i missed shooting an elk at 301yards at -29° I click 3.2MOA.... and missed 5 shot at the elk. I did this math is should be 263yards, and that's .7 MOA.🤔
it's fun to watch him dumb down trigonometry that much. I mean, dude I did go to school lol, but anyways great content!
Best video learned a lot
awesome video as always
Better than any teacher i've ever had in school LOL
Copy and paste cheat sheet if anyone wants it.
5°angle = 0.99 or 99%
10°angle = 0.98 or 98%
20°angle = 0.94 or 94%
30°angle = 0.87 or 87%
40°angle = 0.77 or 77%
50°angle = 0.64 or 64%
60°angle = 0.50 or 50%
70°angle = 0.34 or 34%
80°angle = 0.17 or 17%
90°angle = 0 or 0%
This had been one of the best series I have ever watched. Would have loved to had you as teacher. Thanks so much for such a thorough and simple presentation on a complex and dynamic concept ( shooting ).
If you're shooting downhill can you just look at your barrel and guesstimate what angle it's tilted at in comparison to a straight imaginary line? Great video!
Great explanation! Thanks
Very well taught!
Did you guys notice how a 60 degree angle (not 45 degrees) gives you a .5 Cosine.. I always though a 45 degree would give you half the distance. Glad to learn about these Cosines.
So Ryan, in shooting at an angle, with target downhill (in this example) are you saying you simply substitute your mil height with mil width in the "Mil Estimation" formula, right from the start and then what step would that allow you to skip, the "Angle Range" formula.
I've tuned into your videos time and time again, you have an impressive ability in delivering concise and clear subject matter. Great videos.
Very nice video. One thing tho... What about if I'm shooting uphill ? I understand from your logic that the bullet is being pull to earth harder so the true range is longer than the actual range. The question is : where can I find a chart with cosine for both uphill and downhill angles? I guess the cosine for uphill will start from 1.01 and beyond. I do have a rangefinder that has a built in inclineometer that shows the angle in degrees. Thank you for the videos and taking the time to respond.
Excellent explanation. I wonder what other phenomena affect this corrected distance. For example change in velocity and orientation of bullet?
I wish you would have shown us how to work the formula using the MIL width.
Hi im sory i want to ask Airguns have to find a 180 degree point
for the barrel
If my teachers in HS would have explained tan cos and sin the way he has I would have understood it completely!!! They just tell you the what without showing you the why and drawing a picture and explaining it in real life terms. This guy should make math videos for schools!!! He could keep so many kids from just being lost!!
It would seem that circles (tires) will always be seen on angle as an ellipse, with the greatest length indicating its diameter. Then it remains to the shooter to decide what vehicles have what diameter tires.
Ryan, you make it simple. Thanks!
Btw, how do you like the Mildot analog calculator?
Thank you for this video.
Ryan: I don't want to confuse you here.
Me: Too late.
much better explanation than "hold balls" which is what we were basically taught
a2 + b2 = c2. you are shooting down the hypotenuse of a triangle. if you know your height above a target ( a2 ) and the distance (c2) to the target you can figure out what the bullet thinks the distance is to the target (b2).
It's easier to do COS because you don't need to find your elevation in relation to your target.
Loving your videos bro! you should strike out on your own?!?!? I just finished watching you "understanding MOA" and it was an eye opener for me as i have "never" fired a gun that had a scope on it and just recently purchased a Centerpoint 4x16x44 and still have to mount it.Which i will go back and watch your video on that as well!? :>) Nothings better then a fellow Vet showing people how to shoot.I was in the Navy and only had to qualify with a .22Cal 1911 as well as a .22Cal rifle.As a teen i use to hunt/trap shoot and loved it but after school when i entered the service my life got very busy.I never picked up a firearm again until recently.bout 7yrs ago i decided after a few home invasions in my area that we needed a firearm in the house for our own protection.It didnt stop there though i purchase a bolt gun soon after .I purchased a Savage Axis in 6.5Creedmoor and it has very quickly become my favorite cartridge!:>)Although soon after i have started collecting ARs and their variants! I am up to 5 in total.Two M4s which are my fav. Model!!! But i also have a .224Valkrie and one in 7.62x39 and another in 6.5Grendel.If you cant tell by my purchases i am very much into long range shooting!Im looking to as soon as possible to make a shot at 1mile!!!!!! But in my bluest state of N.J. there wasnt much hope for that.:>/ but just two days we had legislation past that makes my county in N.J. a 2A sanctuary!!!:>) so my hopes have risen!!!:>)I still have to travel out of state to shoot further then 300 or 400yrds!:>/ I thanks you for your content and info i'm grateful! Also i want to thank you for your service!