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Old 11-20-2017, 05:25 AM   #1
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Nuke Torpedo Firing Solution (how germans did it during wwII)

Torpedo Firing Solution
Gyro Angled Shots
(how germans did it during wwII)

These pages will show you how the germans solved the Torpedo Firing problem for gyro-angled shots during wwII.

This pages sequence is devoted to Maciek Florek (aka ''Snakedoc'',subsim nickname ''snakedocpl'') who has left us so early. Maciek had analysed ,at his site, the way Torpedo Vorhalt-Rechner's components (TVR) worked and this helped me very much to understand the solution to firing problem as germans did it at wwII.
(Today , is more easy to solve the Torpedo Firing Problem with geometry-trigonometry on modern computers which 'locking' when a boolean is satisfied. During wwII, were availiable only some mechanicals computers for solving basic equations and therefore they developed the TVR which main purpose was to calculate the gyro angle for the shot).
I want to thank our member ''La vache'' for a document he sent me which ,also, was very helpful.

So, here is how they did it back then (some maths are necessary to understand it):

The most important part to understand is the position of Equivalent Point of Fire (EPF). EPF is the imaginary point that the torpedo tube should have been in order to perform a straight shot to the target at the moment of firing (This point is point N at the drawings). Its position is altered by the demand that its distance from impact point must be equal with the torpedo's travel distance to impact point.The positions of the EPF for every gyro angle could be calculated as explained at the pages and its position for rounded gyro angle values from 0 to 90 can be seen at appendix 1 (page 4).
Next important part to understand is the point N'. N' is the position ,at a specific gyro angle p, that this N should have been in order the 'velocity' triangle to be the correct one.This means that,based on torpedo's and target's speeds, N' is the position that the N should have been,at this specific gyro angle p, in order target and torpedo 'meet' at the impact point.

So , for each gyro angle , two points were altered ,the N (from the 'distance' triangle) and the N' (from the 'velocity' triangle). The 'key' for the Torpedo Firing Solution was to be found that gyro angle which made both N and N' get identical (N≡N').

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