1 mile to 11 feet conversation for dfat/ iota

[4066.5]

Not even sure where this post belongs so feel free to delete it and tell me where but this seems the best place for my dumbass.


For all you math wizards out there. What formula would I use to convert a 24 inch Target at one mile to a paper at 11 feet... Basically I want to know how big to make my target on a sheet of paper at 11-12 feet with my dfat/iota to make to look the same in my reticle as if I'm looking at it at 1 mile/1k/800 ect.

 

[Fig]

The problem is how to figure out how to make it appear equivalent with a variable power scope. We can probably figure out the ratio as a function of distance and angle, but you need to give your lowest power to make it appear that way, and you're probably going to be at max magnification @ 1mi to see a 24" target. I'm trying to figure out how you would express that? Just as a function of the power of the scope as a multiplier?

Not sure of this is a "stupid" marksmanship question, or a smart brain teaser. This from a word guy. I suck at math.

 

[Waorani]

0.05"

 

[4066.5]

The problem is how to figure out how to make it appear equivalent with a variable power scope. We can probably figure out the ratio as a function of distance and angle, but you need to give your lowest power to make it appear that way, and you're probably going to be at max magnification @ 1mi to see a 24" target. I'm trying to figure out how you would express that? Just as a function of the power of the scope as a multiplier?

Not sure of this is a "stupid" marksmanship question, or a smart brain teaser. This from a word guy. I suck at math.

Shit,well, for this purpose it would be through my 27x ffp , but I would like to know what it would be throught my 21x. Does FFP matter ?... I'm thinking about just taking a picture of a 24" target at one mile without zoom.. would that work ?

 

[Fig]

Just as a function of raw angle it would be less than an inch and a half at 100 yards, so @Waorani is either a good guesser, or in the neighborhood @ 11' using maths I do not possess. Teeny tiny at least!

Again, the difficulty is when we're discussing how it appears rather than just being a function of simple math.

I think FFP does matter in terms of how it looks through a reticle that does not stay constant. Maybe @koshkin can shed light. I'm sort of stumped.

 

[Sheldon N]

A 24" target at 1 mile is roughly 0.4 mil wide. Use your FFP reticle and make a target that looks 0.4 mil wide when you are looking at it through your reticle. A piece of sticky paper and a pair of scissors should do the trick.

 

[dirthead1]

Waorani is correct. It's a simple direct proportion. Below is the formula. I've converted everything to feet to make it a little easier.

2' (target) @ 5280' = x' (target) @ 11'

Solve for X

2' / 5280' = X / 11'

Cross multiply to get:

5280 (X) = 22

X=.004166 feet

.004166 feet X 12 to convert to inches = .05"

 

[koshkin]

Just as a function of raw angle it would be less than an inch and a half at 100 yards, so @Waorani is either a good guesser, or in the neighborhood @ 11' using maths I do not possess. Teeny tiny at least!

Again, the difficulty is when we're discussing how it appears rather than just being a function of simple math.

I think FFP does matter in terms of how it looks through a reticle that does not stay constant. Maybe @koshkin can shed light. I'm sort of stumped.

With FFP scopes the reticle technically stays constant with respect to the target, so the relationship of the reticle to the target is the same at all magnifications.

The whole thing looks a little different to your eye because the reticle and the target subtend a different portion of the FOV at different magnifications, but the relationship between reticle and target stays constant.

ILya

 

[Fig]

of course. That’s not what I meant. I meant different sizes depending on the magnification.

 

[koshkin]

of course. That’s not what I meant. I meant different sizes depending on the magnification.

I do not think I understand what you mean. Can you elaborate a little more on this?

ILya

 

[Fig]

From the OP; he wants the target to look the same at 11’ as it does at 1760 yards. The 1/20th of an inch would be with a naked eye. Obviously this is through the scope, so while .4 mils would be constant the apparent size of the target would depend on how much magnification you used. That’s all I meant. Can you just plug the magnification in? I get confused when you add in the relative size of the target through the scope.

 

[koshkin]

From the OP; he wants the target to look the same at 11’ as it does at 1760 yards. The 1/20th of an inch would be with a naked eye. Obviously this is through the scope, so while .4 mils would be constant the apparent size of the target would depend on how much magnification you used. That’s all I meant. Can you just plug the magnification in? I get confused when you add in the relative size of the target through the scope.

If you look at a target at a 24 inch target at 1000 yards through a scope set on 25 magnification, it will look the same (barring any atmospheric effects) as a 2.4 inch target at 100 yards or 0.24 inch target at 10 yards when viewed on the same 25x magnification.

11 feet is too close for most scopes to focus on, so this is a bit of a moot point.

ILya

 

[Mr. Z]

Not with an IOTA or DFAT

 

[4066.5]

Waorani is correct. It's a simple direct proportion. Below is the formula. I've converted everything to feet to make it a little easier.

2' (target) @ 5280' = x' (target) @ 11'

Solve for X

2' / 5280' = X / 11'

Cross multiply to get:

5280 (X) = 22

X=.004166 feet

.004166 feet X 12 to convert to inches = .05"

That's what I was hoping was correct . I'm planning on making more PowerPoint's similar to the one 65guys did but want d a solid starting point.

Thanks !

 

[4066.5]

A 24" target at 1 mile is roughly 0.4 mil wide. Use your FFP reticle and make a target that looks 0.4 mil wide when you are looking at it through your reticle. A piece of sticky paper and a pair of scissors should do the trick.

Ahhhhhh that's so much easier, should have looked at the simpler answer. That will make it easier switching between my moa or mil reticle. As long as a 1 moa target looks like 1 moa at 1k it should be accurate. Trail and error might be the best after I get the math.

 

[acudaowner]

put your 5 shots inside that a 3 oz dixi cup at 100 yards shooting through the open end out the bottom should be close , have seen people shooting who do not have a 1k range shoot 1' circle stickers they say putting 5 either inside the first shot's hole or or touching each other is a good example , I have been trying the stickers cause they are really cheap . so far only 3 touching have I ever been able to do so far but good luck .

 

[4066.5]

View attachment 7095298 put your 5 shots inside that a 3 oz dixi cup at 100 yards shooting through the open end out the bottom should be close , have seen people shooting who do not have a 1k range shoot 1' circle stickers they say putting 5 either inside the first shot's hole or or touching each other is a good example , I have been trying the stickers cause they are really cheap . so far only 3 touching have I ever been able to do so far but good luck .View attachment 7095301

 

[acudaowner]

if this is what is used at 1000 yards
4.7 800, 900, and 1000 Yard Target(a) NRA No. LR - Aiming Black (inches)Rings in White(inches)X ring ....... 10.00 10 ring....... 20.00 9 ring........ 30.00 8 ring........ 44.00 7 ring.......... 60.006 area . . . 72x72 square
than shooting inside a 1' circle at 100 could be used to practice with if the poster of this post did not have 1k range to shoot at

 

[308pirate]

The problem is how to figure out how to make it appear equivalent with a variable power scope. We can probably figure out the ratio as a function of distance and angle, but you need to give your lowest power to make it appear that way, and you're probably going to be at max magnification @ 1mi to see a 24" target. I'm trying to figure out how you would express that? Just as a function of the power of the scope as a multiplier?

Not sure of this is a "stupid" marksmanship question, or a smart brain teaser. This from a word guy. I suck at math.

You do suck at math and are severely confused. The target is reduced proportionally to the distance without any regard to the scope's magnification. That way, the scaling will still be correct regardless of what magnification is used to observe it.

 

[Fig]

I understand that. All I meant (and it IS what I wrote) was that the “apparent” size changes depending on the magnification, so wanting it to “look” the same size in the scope is dependent on the power. Just saying that it’s the equivalent of .05” would be with the naked eye.
I was thinking that 20x at 11’ might not functionally work on a 1/20th” dot. You may have to back off the power and make the dot bigger to be able to actually practice at 11’ as if it were 1000 yds and the target “appears” to be the same size in the scope. Is that more clear?

 

[2aBaC̶a̶]

Interesting, never seen the DFAT thingy.

Im imagining looking through a window sitting 11' away at a 2' object at a mile and it being equivalent to a .05" dot on the window 11' away. About the size of a sharpie dot.

Assuming adjustable power scope would work the same on both.

 

[seansmd]

They have a ppt with scaled targets and their sizes at different ranges at this link, is this what your asking?
http://www.dstprecision.net/index.html near the bottom, practice ranges, not sure how it changes in a different screen size but i believe you can print them 4x7 to get the right scale.

 

[hlee]

The ratio of target height to distance should be equal.

1 mile is 63,360 inches. 11 ft is 132 inches.

12/63,360= 0.000189

0.025/132= 0.000189

0.025 inches. That's really small...