Engravings German WW1 naval binos (1 Viewer)

[Ignatius]

Looking through the internet I have come across several old German naval binoculars from WW1 which have arcane formulae engraved on them.
These have the form E = n x G.
On one of those binoculars, a D.F. 7x50 from 1918 with a FOV of 128 m, n = 7.8 (coincidentally? 1000/128 = 7.8) and on the other one, a 1924 Leitz Aviosept 7x with a FOV of 105 m, n = 9.5 (coincidentally? 1000/105 = 9.5).
Neither of these binoculars has a reticle and therefore I take the formula to be a help for lookouts to determine distances to objects (the German for distance is Entfernung).

I tried using the known length of a warship in metres (filling the image of the binoculars for reference) for G (Grösse?) but that gave me a ridiculously close distance - well within range of naval artillery and almost dangerously close enough for a pie throwing contest.

My question is this: does anyone have any idea what the G in this formula is, and how the relationship of FOV to 1000 m distance might help estimating an actual distance.

 

[Ignatius]

The 1924 Leitz Aviosept

The 1918 Zeiss D.F. 7x50

Both images are in the copyright of Johann Leichtfried's Virtuelles Fernglas-Museum, Austria.

 

[Binastro]

G may be the whole Gross field in metres, not the size of a ship.

So if 5 ships fit in the field, one can work things out.

There probably isn't much distortion with these binoculars.

B.

 

[Ignatius]

So the equation remains one with two unknowns and we are no closer to solving this conundrum.

 

[LPT]

So the equation remains one with two unknowns and we are no closer to solving this conundrum.

This marking on WWI and 1920’s period Leitz Porro II binoculars has baffled collectors for quite a while. It has been discussed on the Binopedia Forum with no resolution, and if anybody would understand what it means, I think somebody there would. For what it’s worth I believe that during the inter-war period Huet made a device which could be attached to one of the objective lenses of a naval binocular to alter magnification and field of view, and I seem to remember reading that Leitz made the same sort of thing. If so, that formula on the objective cover might reference calculating the magnification and/or field of view when such a device is attached.

I don’t think it has anything to do with calculating distance using the size of a warship or other known object. This would be a highly inaccurate way determining distance, and graticules which were routinely installed in binoculars during this time would be a much simpler and more accurate way of doing this.

 

[Ignatius]

I cannot find any trace of any such device by Leitz.
Of course reticles would be the easier way, however, at least these two do not have reticles, so the observer had to work with what he had in hand.
Exactly what the formula MIGHT be is what I am trying to work out. G could be Grad, Geschwindigkeit, Grösse, the letter 'G', ... E could be Entfernung, ...
There must be a way of understanding, what importance the relationship of FOV and 1 km (ie. the decimal numbers) could have, and what the other two unknowns are. Currently I am trawling the internet for Kriegsmarine Dienstanweisungen, equipment manuals etc.
The German navy will not have engraved this out of boredom and the military never make stuff extra difficult because any numpty must be able to deal with this at a pinch, in an emergency and with as little training as possible.

 

[Binastro]

There are not two unknowns.

If one is looking at a ship of known length and if 5.5 ships would fill the field, then the distance is known perhaps to 15% accuracy.

I find that reticles are accurate to about 10%.

There are Nato reticles, maybe 6,400.
Soviet 6,000
Swedish 6,300 but now Nato.

It seems that reticles are not interchangeable.

Nato compasses made in Germany have about six or eight columns and one can read across to get the various equivalents.

Even at my best attempts at FOVs I can only get to about 2% accuracy, maybe 1% with very accurate known star separations.

Firstly, hand held the field moves and persistance of vision makes it impossible to be super accurate.
The eye position is critical.

Add to this there is distortion in binoculars, pincushion, barrel and complex.

So one can only trust the reticles for perhaps 2/3 the field.

Add to this is the magnification might vary by one or two percent.

Lasers are much better, but these also have errors.

Very large baseline optical rangefinders are accurate enough for their purpose, but these might be 20 ft long.

Even using triangulation with simple telescopes I only get about 5% accuracy.

Surveying equipment is much better, by Wild and others.

Regards,
B.

 

[Ignatius]

a) How are there not two unknowns in this formula: E = 7.8 x G

b) How are there not two unknowns in this formula: E = 9.5 x G

Neither bino has a reticle. Both were made decades before NATO reticles or indeed lasers.

But thank you for playing.

 

[Binastro]

G is known.

If the ship is 150m long 5.5 ships are 825 metres field size.

Then E is known.

B.

 

[Binastro]

Rangefinding reticles seem to go back to the 1860s.

Micrometer eyepieces maybe earlier.
With accurate separations and position angle.

B.

 

[Ignatius]

G is known.

If the ship is 150m long 5.5 ships are 825 metres field size.

Then E is known.

B.

I have no clue where you are taking these values from, how you arrive at 5.5 ships and some 'field size', but whatever makes you happy.

Rangefinding reticles seem to go back to the 1860s.

Micrometer eyepieces maybe earlier.
With accurate separations and position angle.

B.

Neither bino has a reticle of any kind, so why harp on about reticles?

The question is about a formula of the form E = n x G with two unknowns (E and G) and a value for n, which for one bino is 7.8 and the other 9.5. No one even seems to know what G is, or indeed E. Which is why I started this thread which sadly seems to be going nowhere fast.

 

[Binastro]

It is very simple.

You estimate the number of ships that will fill the field.

It can be any number.

5.5 ships is just an example.

If the ship's length is known, the full field size can be estimated giving G as a known, but not very exact figure.

Then E, the distance is known.

B.

 

[Binastro]

This does, of course, assume that you are viewing at right angles to the ship.

If this is not so, a correction has to be made by eye estimation.

B.

 

[Ignatius]

I'm mostly talking to myself here, but humour me.

Let us assume E is Entfernung (distance) and G is Gesichtsfeld (FOV).
The formula can be rearranged to solve for G = E/n, and for n = E/G
Let us now assume we are a German lookout cruising along the North Sea in summer 1918, the day is clear. We spot a ship in the distance and our Weyer or Gröner tells us the silhouette is that of HMS XYZ of length 200 m. She takes up 1/4 of the view in our Zeiss D.F. 7x50.
Using the formula we can now estimate that the whole FOV (G) is 4 x 200 = 800. This times 7.8 ( n ) = 6240 (E), ergo the ship is approx. 6.25 km distant.

Could this have been enough in WW1 to set the whole ship's artillery in motion or at least be enough information for deciding on the next steps?

N.B. According to the internet, in WW1 most naval exchanges happened at 16,000 m and less. Also the view was rarely perfect. At a height of 20 m on the bridge, the horizon is 16,000 m away. The smoke could often be seen long before mastheads or funnels even appeared above the horizon.

 

[Binastro]

Eventually some common sense.

A few points though.

Although the horizon at 20m height is 16km, this is the sea horizon.
If looking at a 20m high bridge of another ship the distance it is seen is double this i.e. 32km.
And higher points would be seen also if the air is clear.

The accuracy of the Zeiss binocular is likely to be 10% and at best 5% at the near distance of 6.25km.
At long distances a lot worse.

This is not sufficient for gunnery.

Barr and Stroud introduced the 4.5ft rangefinder in 1892. I have a later version.
I think all Royal Navy ships had these and larger ships had the 9ft version.
Both are accurate to 1%, at 3,000 yards for the 4.5ft and at 7,000 yards for the 9ft.

A 200 metre long ship would have the 9ft version and the second salvo would result in a direct hit.

Using the Zeiss binocular would not be successful.

However, strange as it may be for a land locked country, the Austrian Navy purchased the Barr and Stroud rangefinders as did the Japanese and the U.S.
The U.S then made their own version by Bausch and Lomb.
The Barr and Stroud original was designed by an American inventor.

The 7.8x factor indicates a field of about 7.4 degrees (one radian is 57.3 degrees).

The 9.5x factor indicates a field of about 6 degrees.

B.

 

[Patudo]

A 200 metre long ship would have the 9ft version and the second salvo would result in a direct hit.

If only that were indeed the case. In WW1 British capital ships expended a lot of ammo for, at times, pretty poor results. Of course by then both the firing vessel and the target could be moving at more than 20 knots and ranges exceeded 15,000 yards, meaning that shells were in flight for several seconds. Given the difficulties involved in rangefinding and fire control, I find it pretty amazing that hits were actually achieved at all. They were, but if you look at the actual accuracy of British gunnery in actions like Dogger Bank, Heligoland Bight and Jutland you'll see hit rates were very much less than one every two salvoes.

NB. German rangefinding kit was apparently better in some respects, but their accuracy was not that fantastic either (in absolute terms), though unfortunately still good enough to be very damaging (thanks in part to factors unrelated to optics).

I have no idea what the formula in the original post could possibly be.

 

[Ignatius]

I have written to both Leica and the Zeiss Archives with this, as well as posting the question on a couple more fora. All with Mr. Leichtfried's ok.

 

[Binastro]

Distance in metres equals 1/field in radians times total field of view of binocular in metres at the distance of the ship or known object length in metres at right angles to the viewer.

The example was at 6.25km or 7,000 yards., where the best accuracy was 70 yards for the rangefinder, but probably worse than 350 yards for the binocular.

The problem is the ship wold rarely be at right angles and the length not exactly known, especially with paint to confuse, which was standard practice.
Also movement.

At 15,000 yards accuracy is a lot worse, also because each shot charge would not be exact amongst other things.

However, look at Japan/Russia war where Barr ad Stroud rangefinders were used to good effect.

In WW2 radar controlled guns became very accurate.

For aircraft, wingspans were regularly given to great accuracy which was stupid.
Also the performance of an aircraft was worked out by measuring the diameter of the propeller accurately.

B.