Derivation of the formulas used in the mathematical model of the calculator

The description of the calculator does not contain - due to readability - detailed derivation of all formulas. These derivations are presented below.

Derivation of the formula for calculating the deflection angle

Let's transform the following initial formula:

\[\begin{aligned} sin β = \frac{v_{g}}{v_{t}} * sin γ \end{aligned} \]

in such way, that deflection angle β depends on the torpedo impact angle α and quotient \(\frac{v_{g}}{v_{t}} \).

Because one of the input values of the component for solving the torpedo triangle is quotient \(\frac{v_{g}}{v_{t}} \), let's make following substitution::

\[\begin{aligned} u = \frac{v_{g}}{v_{t}} \end{aligned} \]

which results in:

\[\begin{aligned} sin β = u * sin γ \end{aligned} \]

Because in any triangle, the sum of all three angles is always equal to 180°, the following formula can be written:

\[\begin{aligned} γ = 180° - (α + β) \end{aligned} \]

Having in mind the periodic feature of the sines function – that is the feature that:

\[\begin{aligned} sin α = sin (180° - α) \end{aligned} \]

we get:

\[\begin{aligned} sin γ = sin (α + β) \end{aligned} \]

So finally we get:

\[\begin{aligned} sin β = u * sin (α + β) \end{aligned} \]

This formula can be transformed in the following way:

\[\begin{aligned} sin β = u * (sin α * cos β + cos α * sin β) \end{aligned} \]

\[\begin{aligned} 1 = u * \frac{(sin α * cos β + cos α * sin β)}{sin β} \end{aligned} \]

\[\begin{aligned} 1 = u * (sin α * ctg β + cos α) \end{aligned} \]

\[\begin{aligned} 1 - u * cos α = u * sin α * ctg β \end{aligned} \]

\[\begin{aligned} ctg β = \frac{1 - u * cos α}{u * sin α} = \frac{1}{u * sin α} - ctg α \end{aligned} \]

So the deflection angle β can be calculated based on the quotient of the target and torpedo speed and knowing the torpedo impact angle using the following formula:

\[\begin{aligned} β = arcctg (\frac{1}{u * sin α} - ctg α \end{aligned}) \]

Derivation of the formula for Cartesian coordinates of the equivalent point of fire

The torpedo track consists of three parts: the initial, straight run RT₁,  the arc T₁T₂, when the torpedo turns to its final course, and the segment T₂G, which is the final torpedo run until hitting the target.

Drawing 1

The equivalent point of fire E₁ is located at the line, which overlaps the final torpedo run. It is located in the distance |E₁T₂| equal to the sum of the length of the arc T₁T₂ and the segment RT₁. The length of the segment |RT₁| is constant and for German World War II torpedoes was equal to 9.5 m. The length of the arc |T₁T₂| depended on the gyro angle ρ.

The Cartesian coordinates x₀ and y₀ of the equivalent point of fire E₁ relative to the location of the aiming device Z (which was distanced from the torpedo tubes by 27 meters on type VIIC U-Boats) can be determined in the following way:

The coordinates x_r and y_r of the torpedo tubes R relative to the location of the aiming device Z are:

\[\begin{aligned} (x_{r}; y_{r}) = (27; 0) \end{aligned}\]

The coordinates of the point T₁ relative to the location of the aiming device Z:

\[\begin{aligned} (x_{T1}; y_{T1}) = (x_{r}; y_{r}) + (9,5; 0) = (27; 0) + (9,5; 0) \end{aligned}\]

The position of the point T₂ (relative to the point T₁) can be determined from the following formula (after noticing, that the angle constructed at the arc T₁T₂ is equal to the gyro angle ρ):

\[\begin{aligned} (x_{T2}; y_{T2}) = (r * sin ρ; r – r * cos ρ) \end{aligned}\] 

where r – turn radius, which was equal to 95 m for German World War II torpedoes.

Whereas, relative to the aiming device location, the coordinates are:

\[\begin{aligned} (x_{T2}; y_{T2}) = (r * sin ρ; r – r * cos ρ) + (27; 0) + (9,5; 0) \end{aligned}\]

Drawing 2

The length of the arc T₁T₂ is

\[\begin{aligned} |T_{1}T_{2}| = ρ * r \end{aligned}\]

where angle ρ is expressed in radians.

As mentioned before, the distance between the point E₁ and point T₂ is equal to the sum:

\[\begin{aligned} |E_{1}T_{2}| = |RT_{1}| + |T_{1}T_{2}|  \end{aligned}\]

that is

\[\begin{aligned} |E_{1}T_{2}| = 9,5 + ρ * r \end{aligned}\]

Coordinates of the point E₁ relative to the point T₂ are:

\[\begin{aligned} (x_{E1}; y_{E1}) = (x_{T2}; y_{T2}) – (|E_{1}T_{2}| * cos ρ; |E_{1}T_{2}| * sin ρ) \end{aligned}\]

Drawing 3

Coordinates of the point E₁ relative to the aiming device Z location:

\[\begin{aligned} (x_{E1}; y_{E1}) = (r * sin ρ; r – r * cos ρ) + (27; 0) + (9,5; 0) – (|E_{1}T_{2}| * cos ρ; |E_{1}T_{2}| * sin ρ) \end{aligned}\]

After inserting the formula determining the length of the segment |E₁T₂| we get:

\[\begin{aligned} (x_{E1}; y_{E1}) = (r * sin ρ; r – r * cos ρ) + (27; 0) + (9,5; 0) – ((9,5 + ρ * r) * cos ρ; (9,5 + ρ * r )* sin ρ) \end{aligned}\]

So the coordinates of the point E₁ are:

\[\begin{cases} x_{E1} = r * sin ρ  + 27 + 9,5 – (9,5 + ρ * r) * cos ρ \\ y_{E1} = r – r * cos ρ – (9,5 + ρ * r) * sin ρ \end{cases}\]

that is

\[\begin{cases} x_{E1} = 27 + 9,5 + 95 * sin ρ  – (9,5 + ρ * 95) * cos ρ \\ y_{E1} = 95 * (1 – cos ρ) – (9,5 + ρ * 95) * sin ρ \end{cases}\]

Derivation of the formula for calculating the torpedo salvo spread angle

The transformation of the following formula

\[\begin{aligned} \frac{E - λ * cos γ}{v_{t} * cos β + v_{g} * cos γ} = \frac{λ * sin γ}{v_{t} * sin β – v_{g} * sin γ} \end{aligned}\]

we are beginning from dividing denominators at both sides of equation by v_t, which results in

\[\begin{aligned} \frac{E - λ * cos γ}{cos β + \frac{v_{g}}{v_{t}} * cos γ} = \frac{λ * sin γ}{sin β – \frac{v_{g}}{v_{t}} * sin γ} \end{aligned}\]

Then the following substitution is done:

\[\begin{aligned} u = \frac{v_{g}}{v_{t}} \end{aligned}\]

getting:

\[\begin{aligned} \frac{E - λ * cos γ}{cos β + u * cos γ} = \frac{λ * sin γ}{sin β – u * sin γ} \end{aligned}\] 

After dividing the numerators at both sides of equation by E we get:

\[\begin{aligned} \frac{1 - \frac{λ}{E} * cos γ}{cos β + u * cos γ} = \frac{\frac{λ}{E} * sin γ}{sin β – u * sin γ} \end{aligned}\]

Then we make following substitution:

\[\begin{aligned} μ = \frac{λ}{E} \end{aligned}\]

getting:

\[\begin{aligned} \frac{1 - μ * cos γ}{cos β + u * cos γ} = \frac{μ * sin γ}{sin β – u * sin γ} \end{aligned}\]

Then the following transformations are performed:

\[\begin{aligned} (1 - μ * cos γ)*(sin β – u * sin γ) = (μ * sin γ)*(cos β + u * cos γ) \end{aligned}\]

\[\begin{aligned} sin β - u * sin γ - μ * cos γ * sin β + μ * u * sin γ * cos γ = μ * sin γ * cos β + μ * u * sin γ * cos γ \end{aligned}\]

\[\begin{aligned} sin β - u * sin γ = μ * sin γ * cos β + μ * cos γ * sin β \end{aligned}\]

\[\begin{aligned} sin β - u * sin γ = μ * (sin γ * cos β + cos γ * sin β) \end{aligned}\]

finally getting:

\[\begin{aligned} sin β - u * sin γ = μ * sin (γ + β) \end{aligned}\]

This is implicit equation, which can be used for calculating the deflection angle β for the target as seen with the angle on the bow γ, so the torpedo hits the point distanced by the value λ from the middle of the target.

For convenience, instead considering the stern and bow location as distances λ from the middle of the target, the following substitution can be done:

\[\begin{aligned} μ_{0} = \frac{l}{2*E} \end{aligned}\]

where l – length of the target.

To make three-torpedo salvo hitting the bow, middle and stern of the target of length l, each torpedo has to be launched with the deflection angle β₀, β and β₁ respectively, where each of these values meets one of the following equations:

\[\begin{cases}sin β_{0} – u * sin γ = μ_{0} * sin (β_{0} + γ) \\ sin β – u * sin γ = 0  \\ sin β_{1} – u * sin γ = - μ_{0} * sin (β_{1} + γ)\end{cases}\]

Because, in practice, in most cases the target (with the length up to 150 meters – i.e. Liberty type cargo ships: ~135 meters, T2 type tankers: ~152 meters) were attacked from the minimum distance 500 meters, the value μ₀ is rather small (μ₀ < 0,15). That means, that angle between the courses of torpedoes aimed into target bow and stern (that is angle  β₁ - β₀) is equal maximum ~20º. Knowing the fact, that for the angles α in range 0 – 20º, the sin α ~ α and cos α ~ 1, the following equations can be written:

\[\begin{align} sin β_{0} = sin (β_{0} - β + β) = sin (β_{0} - β) * cos β + cos (β_{0} - β) * sin β \end{align}\]

and

\[\begin{align} sin β_{1} = sin (β_{1} - β + β) = sin (β_{1} - β) * cos β + cos (β_{1} - β) * sin β \end{align}\]

Using the feature of the sines and cosines functions for small angles mentioned above, the equations above can be written in the following way:

\[\begin{align} sin β_{0} = (β_{0} - β) * cos β + sin β \end{align}\]

and

\[\begin{align} sin β_{1} = (β_{1} - β) * cos β + sin β \end{align}\]

After inserting the formulas above into previous equations, we get:

\[\begin{cases} (β_{0} - β) * cos β + sin β - u * sin γ = μ_{0} * sin (β_{0} + γ) \\ sin β - u * sin γ = 0 \\ (β_{1} - β) * cos β + sin β - u * sin γ = -μ_{0} * sin (β_{1} + γ) \end{cases}\]

and after transforming:

\[\begin{cases} sin β - u * sin γ = μ_{0} * sin (β_{0} + γ) - (β_{0} - β) * cos β \\ sin β - u * sin γ = 0 \\ sin β - u * sin γ = -μ_{0} * sin (β_{1} + γ) - (β_{1} - β) * cos β \end{cases}\]

After inserting the second equation into the two others, we receive two formulas:

\[\begin{cases} 0 = μ_{0} * sin (β_{0} + γ) - (β_{0} - β) * cos β \\ 0 = -μ_{0} * sin (β_{1} + γ) - (β_{1} - β) * cos β \end{cases}\]

By comparing these two equations, we get:

\[\begin{aligned} μ_{0} * sin (β_{0} + γ) - (β_{0} - β) * cos β = -μ_{0} * sin (β_{1} + γ) - (β_{1} - β) * cos β \end{aligned}\]

which results in:

\[\begin{aligned} β_{0} - β_{1} = μ_{0} * \frac{sin (β_{0} + γ) + sin (β_{1} + γ)}{cos β} \end{aligned}\] 

As mentioned before, the difference of the angles β₁ - β₀  in practice are maximum 20º. The sines function for angle values up to 20º can be approximated by linear function. Se we can write:

\[\begin{aligned} sin (β + γ) = \frac{sin (β_{0} + γ) + sin (β_{1} + γ)}{2} \end{aligned}\]

while

\[\begin{aligned} β \approx \frac{β_{0} + β_{1}}{2} \end{aligned}\]

So the torpedo salvo spread angle ψ (which is the difference β₁ – β₀) can be calculated by following formula:

\[\begin{aligned} ψ = 2 * μ_{0} * \frac{sin (β + γ)}{cos β} \end{aligned}\]

This equation can be transformed in the following steps, to avoid dependency from the deflection angle β:

\[\begin{aligned} ψ = 2 * μ_{0} * \frac{sin (β + γ)}{cos β} = \frac{l}{E} * \frac{sin (β + γ)}{cos β} = \frac{l}{E} * \frac{sin β * cos γ + cos β * sin γ}{cos β}\end{aligned}\]

\[\begin{aligned} ψ = \frac{l}{E} * \bigg(\frac{sin β * cos γ}{cos β} + \frac{cos β * sin γ}{cos β}\bigg) = \frac{l}{E} * (tg β * cos γ + sin γ)\end{aligned}\]

\[\begin{aligned} ψ = \frac{l}{E} * sin γ * \bigg(\frac{tg β * cos γ}{sin γ} + 1\bigg) = \frac{l}{E} * sin γ * \bigg(\frac{cos γ}{ctg β * sin γ} + 1\bigg) \end{aligned}\]

\[\begin{aligned} ψ = \frac{l}{E} * sin γ * \bigg(\frac{cos γ}{\frac{cos β}{sin β} * sin γ} + 1\bigg) = \frac{l}{E} * sin γ * \bigg(\frac{cos γ}{\sqrt{\frac{cos^2 β}{sin^2 β} * sin^2 γ}} + 1\bigg) \end{aligned}\]

\[\begin{aligned} ψ = \frac{l}{E} * sin γ * \bigg(\frac{cos γ}{\sqrt{\frac{1 - sin^2 β}{sin^2 β} * sin^2 γ}} + 1\bigg) = \frac{l}{E} * sin γ * \bigg(\frac{cos γ}{\sqrt{\frac{sin^2 γ - sin^2 β * sin^2 γ}{sin^2 β}}} + 1\bigg) \end{aligned}\]

\[\begin{aligned} ψ = \frac{l}{E} * sin γ * \bigg(\frac{cos γ}{\sqrt{\frac{sin^2 γ}{sin^2 β} - \frac{sin^2 β * sin^2 γ}{sin^2 β}}} + 1\bigg) \end{aligned}\]

Because

\[\begin{aligned} \frac{sin γ}{sin β} = \frac{v_{t}}{v_{g}} \end{aligned}\]

finally we get:

\[\begin{aligned} ψ = \frac{l}{E} * sin γ * \bigg(\frac{cos γ}{\sqrt{(\frac{v_{t}}{v_{g}})^2 - sin^2 γ}} + 1\bigg) \end{aligned}\]