Scales and real-thing equivalence

Maybe a simpler, less dangerous pursuit would be to use a 1/24th scale car kit that is 1/24th the length of your own car (doesn't have to be an exact model of your car) - viewed at measurable distances?

WmB

You might be able to figure out something in the comfort of your own home. I grabbed the closest thing I could find which turns out to be a 3x5 index card. I cut it in half, took one of the halves and cut it down a few more times so that it is 1/4 the size (by area) of the first half - 1/2 scale.

I can hold these two pieces up, moving the larger of the two further away until they are both the same size. If one were so inclined and not headed to bed shortly ;-), you could measure the distances and scales and fiddle around with it - taking into acct parallax and eye dominance I suppose.

That might be a quick and dirty desktop experiment that could help corroborate your figures.

WmB

Yes, that's a good one. Meanwhile, enormous 1:1 scale experiments similar to those that you describe but in my back garden have led me to the conclusion that my initial estimates are crap, I think because I corrected twice for the arm's-length bit. Here are (what I think are) some better estimates:

Viewing a 1/24th scale kit from 3 feet away is equivalent to viewing the real thing from about 15 yards

Viewing a 1/32nd scale kit from 3 feet away is equivalent to viewing the real thing from about 19 yards

Viewing a 1/48th scale kit from 3 feet away is equivalent to viewing the real thing from about 25 yards

Viewing a 1/72nd scale kit from 3 feet away is equivalent to viewing the real thing from about 32 yards

Viewing a 1/144th scale kit from 3 feet away is equivalent to viewing the real thing from about 48 yards

Sound about right?

Nick.

sort of off topic, I've always wanted a way to figure out the altitude of the planes flying over my house. They are on approach or just taking off from SFO or Oakland. I just make stuff up for the kids, well they look to be around 1-2 inches long so that puts the 747 at

7000 feet. Anyone got a way to give them a little more accurate guess??

Craig

Not even close. Inverse-square relationships are for *area*, not linear size. And there's no need to correct for the length of your arm. You're only using your arm to hold the measuring device at a constant distance from your eyes, so that its apparent size remains constant. All that matters is the distance of the object being measured from your eyes.

As for the correct numbers, the apparent size of a model kit is just a direct inverse relationship. Viewing a 1/24 scale kit from 3 feet is equivalent to viewing the real thing from (3x24) = 72 feet = 24 yards. Viewing a 1/144 scale kit from 3 feet is equivalent to viewing the real thing from (3x144) feet = 144 yards.

Ask any physicist, or any artist who understands perspective. He'll tell you that apparent size is in direct inverse proportion to the distance.

Here's a simple experiment: Take three sticks or boards 2, 4, and 6 feet long. Lay them on the ground, spaced 10 feet apart and parallel, like this:

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Sit or lie down 10 feet from the shortest one, and look at them. The 4' stick should be twice as far from you as the 2' stick, and the 6' stick should be three times as far as the 2' stick, but they'll all appear to be the same length.

You can also do this with boxes of different sizes, or with two models of the same subject in different scales (e.g. 1/24 and 1/48). If the bigger object is 1.5 times longer than the small one, place it 1.5 times farther away from you and they'll appear to be the same size.

In simple terms, viewed from 3 feet away, the equivalent viewing distance in yards is the denominator of the scale, yes?

1/24 = 24 yards 1/32 = 32 yards 1/48 = 48 yards

etc.

BobbyG

I remember reading somewhere about why 1:72 was chosen for the identification models used in WW II - something about the equivalent size of the real aircraft viewed at a given distance when the model was held at arms length. Anybody know more about this?

Jack G.

Arm length is not quite a meter. So distance would be around 60 to 70 meters. Instructor: "If you see the plane in this size, it is too late to shoot at him." :-)

Wayne & Bobby are right with their math.

Nick

Bingo. I tried my little paper experiment a little while ago and here's what I found:

A 1/2 scale paper square placed 1/2 the distance between the full paper square and the eye appear the same size. A 1/4 scale object placed 1/4th of the distance between eye and FS object appears the same size. I tried this by varying the distances between the eye, the full sized object and the scaled down object and it came up close enough for govt work each time.

So it's the inverse like Wayne said.

L1 - The distance between the full scale object and your eye L2 - The distance between the scaled down model and your eye SF = Scale factor (fractional expression)

L2= L1*SF

Ex: For L2 = 1 yard, SF=1/24, the distance to the full size object, L1 =

1/(1/24) = 24 yards (72 ft)

WmB

Thanks Wayne -- good points. It's the area (which corresponds to the size of the 2D image on your retina) that will follow an inverse square rule, but any straight-line drawn through that will be decreasing in a direct inverse way, as you've outlined.

I've discovered that classical perspective is still not the full story when it comes to what we perceive as a 3D object. Apparently binocular vision operates in a non-linear way. I found a good article here:

"Each of your eyes, like a camera lens, sees the world in "correct" classical perspective. Since binocular vision takes our perception of space to a stage beyond what we can see with just one eye, it follows that it operates by exaggerating and enhacing perspective in generating a single new "3-D" image out of the two "2-D" images the brain receives and transforms. Binocular depth perception might be described mathematically through some kind of non-Euclidian geometry..."

After that, there's also the psychological thing about things on the horizon plane appearing to be much larger than those in the sky. This one's called the "moon illusion" Everyone seems to agree that it exists, but there still seems to be debate about it cause. That one is really worth reaing about -- here's a link:

Thanks again, Nick.

I'm interested in the answer to this too. We know the size of a 747 (231 ft 10 in), and the size you see it (2 inches), and can work out that that you're seeing it 1391 times smaller than the original. But what now?

I guess you could pretend it's a 1:1391 scale model, and say to yourself: viewing a 1391 scale model at 3 feet (you're holding out a ruler in front of your face) is the same as seeing the real thing at

4173 yards.

(This is where the distance of the ruler from your eyes has to matter, because the scale of a ruler becomes gigantic just before you lose focus!)

Or in metric, with your ruler at 1 m, that would imply the thing is

1391 metres away (same answer, but different units).

I wonder if that's right?

Method 2:

If you had a really precise protractor, I think you could do it by a variation of Ptolemy's method for measuring the height of buildings.

Do you think you'll be able to bluff your kids into running around the back yard with protractors?

Ptolemy reiled on the existence of a right-angle between the ground and the building. The lines joining each end of the plane to your eye also form a triangle -- if you cut that in half in the middle of the plane as it goes overhead, you've got two identical right-angled triangles.

Once you measure the total angle of the 747 from back to front as it passes overhead, the equation for working out the distance is:

Distance = 0.5real size / tan(angle/2)

The units can be feet or metres or whatever, as long as they are the same for distance as the size of the aircraft.

Half the real size of a 747 is 35.335 metres.

So if the angle you saw from front to back of the 747 was 1 degree, then tan(1/2) = 0.00872688, and the implied distance is 35.335/0.0087 = about 4000 metres.

Coming in to land it blocks out a 45 degree arc of sky as it passes overhead. That would put its distance at 35.335 m / tan(45/2) or 85 metres.

A warning though! This equation STRONGLY suggests that once the angle on your protractor hits 180 degrees, you have ALREADY been squashed under a 747!

Nick.

Nick,

In short, Wayne and WmB are right: a 1/N-scale model viewed at one yard looks the same as the real thing viewed at N yards. Moreover, that's where you should stop worrying about the subject, in my opinion.

Although you're correct in noting that all of the earlier discussion on this thread has ignored stereoscopic cues, binocular depth perception can't be accounted for in a diorama (which I infer is what you have in mind) without requiring the observer to view the diorama from a fixed position through a sophisticated optical system that changes the effective distance between his/her eyes -- which probably isn't practical for you.

As for the "moon illusion," I doubt strongly that it's at all relevant to a diorama, which *always* will include the "distance-cues" that Professor McCready mentions in Section III of his article, under 'Eye Adjustments in the Moon Illusion.'

Charles Metz, who has some professional experience in this area

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Nick Kim wrote:

I don't follow your math. Viewing a 1/24 scale model at 3 feet (1 yard) is equivalent to viewing te real thing from 24 yards. A 1/32 scale kit from 32 yards, etc.

Nick Kim wrote:

True; unfortunately, as anybody who's really tried to make realistic looking dioramas or photos finds out, you can't just scale things linearly down and expect them to work; they always look like models. You need to jigger all kinds of things, like scale of background objects, etc. It's an art not a science, and that means I'm hopeless at it.