Computing Gun Impulse/Recoil
Forces

Cutting Down the Kick: Understanding and Managing Large Caliber Recoil, April 2015 (2.8~ MB PDF)
The Designer’s Dilemma – Recoil: What to do with it? (RTOMPAVT108) (768~ KB PDF)
A Fire Out of Battery Tank Gun: Theory and Simulation (829~ kb PDF)
Design Tradeoffs For A Very Lightweight 155mm Howitzer For The U.S. Army Light Forces (1.7~ MB PDF)
AMCP 706251 Engineering Design Handbook Guns Series, Muzzle Devices (May 1968) (5.7~ MB PDF)
Reducing the average recoil force via increasing the recoil distance lowers the stresses on the turret structure and the rest of the vehicle, but it does not reduce the firing impulse – and it is the firing impulse that governs whether a gun can be put in a vehicle without suffering:
Excessive movement during firing.
High risk of overturning when the gun is fired over the side.
Undue discomfort to the crew and mechanical components when firing.
As a rule of thumb, the comfortable firing impulse for a vehicle mounted weapon is generally held to be between 0.5 to 0.7 kNs/tonne. The Ogorkiewicz Limit is 0.9 kNs/tonne. Anything beyond that will be extremely uncomfortable for the crew; with a high chance of breaking various components – the M551 Sheridan exceeded 1 kNs/tonne when it fired it's 152mm gun with HEAT rounds.
Firing Impulse can be reduced through the use of muzzle brakes, which create an impulse in the opposite direction via deflecting the flow of propellant gases through the muzzle. Unfortunately, the higher efficiency muzzle brakes gain their efficiency at a price – through tremendous overpressure.
In some of the early tests with the Stryker Mobile Gun System, the overpressure from the muzzle brake was enough to shatter the driver's vision blocks. Additionally, infantry providing close support to the vehicle might also not like it when the gun fires and the muzzle brake can kick up a large amount of dust during firing, obscuring the target in your sights, or providing an easy location marker for the enemy.
Reference: The Designer’s Dilemma – Recoil: What to do with it? (RTOMPAVT108) (768~ KB PDF) 
Reference Gun Impulses 

Weapon 
Weight (kg) 
Impulse (NSec) 
Nsec/kg 
155mm M1917 Howitzer 
3,750 
30,400 
8.11 
155mm M1918M1 Gun 
11,300 
49,660 
4.39 
155mm M1 Gun 
12,700 
57,630 
4.54 
155mm M1/M114 Howitzer 
5,765 
38,170 
6.62 
155mm M198 GunHowitzer 
7,163 
46,260 
6.46 
155mm M777 GunHowitzer 
4,218 
46,260 
10.97 
155mm XM282 Howitzer 
– 
59,508 (without brake) 
– 
Reference: The Designer’s Dilemma – Recoil: What to do with it? (RTOMPAVT108) (768~ KB PDF) 
F_{Recoil} = I^{2} / (2 * M_{Recoil} * D_{Recoil})
Where:
F_{Recoil} = Average force on the gun mounting in Newtons (N)
I = Firing Impulse in Newtonseconds (Ns)
M_{Recoil} = Mass of recoiling parts in kilograms (kg).
D_{Recoil} = Recoil distance in meters (m).
Reference:
Technology of Tanks by Richard M. Ogorkiewicz
EXAMPLE: A modern 105mm gun has a firing impulse of 16,710 NewtonSeconds (Ns), a recoiling mass of 1,350 kg and a recoil distance of 0.280 meters. What is its average recoil force? 16,710^{2} / (2 * 1,350 * 0.280) = 369,344.05 Newtons (369.34 kN) 
I_{Braked} = I * SQRT(1 – E_{Brake})
Where:
I_{Braked} = Firing Impulse on the gun with a muzzle brake in NewtonSeconds (Ns)
I = Firing Impulse on the gun without a muzzle brake in NewtonSeconds (Ns)
E_{Brake} = Efficiency of the Muzzle Brake
Notes: Generally, Muzzle Brake efficiency has been between 2540%, but purpose built lightweight guns have had even higher efficiencies; with the Cockerill 90mm Mk7 having an efficiency of 55%, while the MECAR 90/46 has an efficiency of 70%.
Reference:
Technology of Tanks by Richard M. Ogorkiewicz
EXAMPLE: Our gun has a firing impulse of 50,000 newtons. If we fit it with a muzzle brake 50% efficient, what would it's firing impulse be? 50,000 * SQRT(1 – 0.50) = 35,355 newtons 

I = (M_{Proj} * V_{Proj}) + (M_{Charge} * V_{Charge})
Where
I = Impulse in NewtonSeconds (Ns)
M_{Proj} = Mass of projectile in kilograms (kg).
M_{Charge} = mass of propellant charge in kilograms (kg).
V_{Proj} = Muzzle velocity of projectile in m/sec.
V_{Charge} = Average velocity of propellant gases in m/sec.
Reference:
Technology of Tanks by Richard M. Ogorkiewicz
EXAMPLE: A modern 105mm gun is firing a projectile massing 5.82 kilograms at a muzzle velocity of 1,500 m/sec, with a charge weight of 5.7 kilograms and a charge velocity of 1,400 m/sec. What is it's impulse force? (5.82 * 1500) + (5.7 * 1400) = 16,710 NewtonSeconds (16.7 kNs) 
V_{Charge} = V_{Proj }/ β
Where
V_{Charge} = Average velocity of propellant gases in m/sec.
V_{Proj} = Muzzle velocity of projectile in m/sec.
β = Coefficient. Varies according to muzzle velocity. Known coefficients according to the Rheinmetall Handbook of Weaponry are:
400 m/sec = 3
800 m/sec = 1.5
1,400 m/sec = 1.0
Extrapolating a powerlaw based on those three points gives us the following equation for β:
β = 576.17 * V_{proj}^{0.877}
and the following coefficients:
100 m/sec = 10.15
200 m/sec = 5.52
300 m/sec = 3.87
400 m/sec = 3
500 m/sec = 2.47
600 m/sec = 2.1
700 m/sec = 1.84
800 m/sec = 1.63
900 m/sec = 1.47
1000 m/sec = 1.34
1100 m/sec = 1.23
1200 m/sec = 1.14
1300 m/sec = 1.07
1400 m/sec = 1
1500 m/sec = 0.94
1600 m/sec = 0.89
1700 m/sec = 0.84
1800 m/sec = 0.8
1900 m/sec = 0.76
I = (Mass_{Proj} * Velocity_{Muzzle}) + (Mass_{Charge} * (Velocity_{Muzzle }* β))
Where
I = Impulse in Newtons
Mass_{Proj} = Mass of projectile in kilograms
Mass_{Charge} = mass of propellant charge in kilograms
Velocity_{Muzzle} = Muzzle velocity of projectile in m/sec.
β = Coefficient. Varies according to muzzle velocity. Typical Coefficients are:
400 m/sec = 3
800 m/sec = 1.5
1,400 m/sec = 1.0
Reference:
Rheinmetall Handbook of Weaponry
EXAMPLE: A modern 120mm gun is firing a projectile massing 7 kilograms at a muzzle velocity of 1,400 m/sec, with a charge weight of 10 kilograms. What is it's impulse force? (7 * 1400) + (10 * (1400 * 1.0)) = 23,800 newtons 