There has been a lot of arguments over the performance of A2A missiles here --

some of which are well backed by facts, some are so far out there it is like claiming that

they fly on hyperdrive.

I want to take this opportunity to introduce everyone to a very simple formula that

can be used for estimating the performance of a missile. It goes like this:-

Change in Velocity (Delta V) = 10 x Specific Impulse x LN (initial weight / final weight) m/s

This assumes that all the fuel is used to get the missile as fast as possible and

none is used to provide just enough thrust to sustain a given velocity.

In otherwords, it assumes an all-boost motor not a boost sustain motor.

For example, let'a take a look at the AIM-120A AMRAAM which we have some decent info on...

Launch weight = 335 lbs (Published stats)

Motor weight = 156 lbs (WPU-6/B HTPB rocket motor weight as per Raytheon)

Approximate specific impulse = 245 seconds (typical of HTPB solid motors)

Approximate fuel fraction of motor = 85% (typical of robust aluminum cased aerospace rocket motors)

OK... if 85% of the motor's mass is the fuel, we have about 132 lbs of fuel in the AMRAAM-A

-- roughly a 39.4% fuel fraction (sounds about right). So let's run the numbers...

Delta V = 10 x 245 x LN(335/(335-132)) = 1227 m/s

The formula predicts that the AMRAAM will go about 1227 m/s (~Mach 3.7) faster than it started.

If it is launched at say Mach 1.5 it'll be going Mach 5.2.

In reality the AMRAAM doesn't go that fast.

The reason is that not all the fuel is used to get it as fast as possible.

The AMRAAM's motor is a boost-sustain design.

It is probably grained to take the weapon to abut Mach 2.5~2.8 faster than it started at

(Mach 4+ in a typical Mach 1.5 release).

The rest of the fuel is shaped to burn much more slowly to keep it's velocity at

or near the achieved maximum out to a longer range before the motor burns out.

Well, for any given fuel fraction and specific impulse,

a designer can decide how fast he wants to go and how long he wants to stay at

or near the peak velocity achieved. For instance, if a missile carries 40% of its launch weight

as fuel and uses the typical a modern HTPB propellant motor, it can:-

(1) Spend 25% to get an approximate Mach 2.1 delta V and 15% on sustaining that speed for a relatively long while.

(2) Spend 30% to get an approximate Mach 2.7 delta V and 10% on sustaining that speed for a shorter while.

(3) Spend 40% to get an approximate Mach 3.8 delta V have no sustain burn time at all.

BTW, in reference to the above comment on deceleration... it doesn't really work that way.

If a missle starts at Mach 4 at burn out and decelerates 25% to Mach 3 after 10~15 seconds,

it WILL NOT decelerate to Mach 2 (another 33% from Mach 3) after 20~30 seconds.

This is impossible because aerodynamic drag (Fd = Cd x A x 0.5 x P x V^2) is a function of

the square of velocity.

As velocity decreases, drag force decreases exponentially in relation to it.

Hence, if the drag for at Mach 4 causes a 25% loss in velocity in 10~15 seconds,

there is no way a much lower drag force at Mach 3 will cause a 33% loss in velocity after

another 10~15 seconds.

What happens is that deceleration is non-linear;

you start off steep and the slope flattens out over time as velocity and hence drag drops.

It'll take a missile a heck of a lot longer to decelerate from Mach 4 to Mach 2 compared to

say Mach 2 to Mach 1 for instance.

Actually it also depends a heck of a lot on altitude (air density)...

Let's plug some numbers shall we?

Question: How much thrust is needed to sustain Mach 3.0 in an AAM like the AMRAAM?

Drag force (Newtons) = 0.5 x P x V^2 x Cd x A

P = Density of Air (kg/m^3) ; ~1.29 kg/m^3 @ sea level; ~0.232 kg/m^3 @ 12,000 m

V = Velocity (m/s) ; Mach 1 = 340 m/s @ sea level; ~295 m/s @ 12,000 m

Cd = Co-efficient of Drag ; ~ 0.6 to 0.95 for rockets depending mostly on finnage,

nose and tail profile

A = Sectional Area (m^2) ; ~ 0.025 m^2 for a 7" diameter missile.

For an AMRAAM like AAM going at high altitudes (40,000 ft)...

Drag Force @ Mach 3 = 0.5 x 0.232 x (295x3)^2 x 0.70 x 0.025 = 1590 Newtons = 357 lbs

Drag Force @ Mach 2 = 0.5 x 0.232 x (295x2)^2 x 0.70 x 0.025 = 707 Newtons = 159 lbs

Drag Force @ Mach 1 = 0.5 x 0.232 x 295^2 x 0.70 x 0.025 = 177 Newtons = 39.8 lbs

The same missile going Mach 3 at Sea Level...

Drag Force @ Mach 3 = 0.5 x 1.29 x (340x3)^2 x 0.70 x 0.025 = 11,744 Newtons = 2640 lbs

Drag Force @ Mach 2 = 0.5 x 1.29 x (340x2)^2 x 0.70 x 0.025 = 5,219 Newtons = 1173 lbs

Drag Force @ Mach 1 = 0.5 x 1.29 x 340^2 x 0.70 x 0.025 = 1,305 Newtons = 293 lbs

Assuming that there is no sustainer,

the deceleration experienced at Mach 3 by the 203 lbs (empty) missile is

Deceleration @ Mach 3 = -F / mass = -1590 / (203 x 0.454) = -17.3 m/s^2 = - Mach 0.059/sec @ 40,000 ft

Deceleration @ Mach 2 = -F / mass = -707 / (203 x 0.454) = -7.67 m/s^2 = - Mach 0.026/sec @ 40,000 ft

Deceleration @ Mach 1 = -F / mass = -177 / (203 x 0.454) = -1.92 m/s^2 = - Mach 0.0065/sec @ 40,000 ft

Deceleration @ Mach 3 = -F / mass = -11744 / (203 x 0.454) = -127 m/s^2 = - Mach 0.39/sec @ sea level

Deceleration @ Mach 2 = -F / mass = -5219 / (203 x 0.454) = -56.6 m/s^2 = - Mach 0.17/sec @ sea level

Deceleration @ Mach 1 = -F / mass = -1305 / (203 x 0.454) = -14.2 m/s^2 = - Mach 0.042/sec @ sea level

OK... enough of the math and the formulas... what does all these mean?

Well, it means that while coasting at Mach 3 an AAM is going to lose about less than 2% of

its velocity a second at high altitudes while it stands to lose about 13% of its velocity at

sea level! Huge difference isn't it?

Remember though that the rate of deceleration actually DECREASES as the

missile's velocity decreases.

It is easy to see that one can claim that a missile can burn out burn out its booster

and sustainer and be effective out to over 100 km at high altitudes or be useful only

against helos after 10km on the deck!

Also, we can make a pretty educated guess as to how much thrust the sustainer has to make.

An AMRAAM class missile with a 400 lbs sustain thrust will be able to stay

above Mach 3 at high altitudes and stay about Mach 1.2 at sea level.

An AMRAAM class missile carrying about 10% of its launch weight as sustainer

grained propellant will be able to keep this level of thrust lit for 20.5 seconds

in addition to whatever the boost time was using the 30% of its fuel to get a

roughly Mach 2.7 Delta V after launch.

A missile like this when fired at Mach 1.5 will reach Mach 4+ and keep

above Mach 3 for the duration of the sustainer at high altitudes.

It will also reach about Mach 2.5 and keep above about Mach 1.2 at sea level.

A motor grained for this thrust profile can have a 10 second boost at ~ 2460 lbs thrust and

a 20 second sustain burn at 400 lbs thrust -- this is a 5:1 boost sustain ratio.

This is also about right for thrust profiles of star grain vs

core burn solid propellant burn rate profiles.

Another rough rule of thumb:-

The time it takes for a missile to lose 25% of its velocity after burn out at supersonic speeds.

Never @ > 100,000 m (~300,000 ft) ; in space

~150 seconds @ 24,000 m (~80,000 ft)

~70 seconds @ 18,000 m (~ 60,000 ft)

~25 seconds @ 12,000 m (~ 40,000 ft)

~10 seconds @ 6,000 ft (~20,000 ft)

~5 seconds @ Sea Level

Remember, fractions over time are not additive.

In otherwords, if a missile loses about 25% of its velocity in 10 seconds,

in the 10 subsequent seconds (t =20s) the missile loses approximately another 25% of

the remaining 75% not a 100%. Total velocity loss is ~43.75% not 50%.

This is highly collated to the fall in air density.

Drag = 0.5 x P x V^2 x Cd x A.

Holding everything else constant Drag falls proportionally to density.

Drag also falls exponentially with Velocity which accounts for the loss in velocity

in the given time slices being about 25% instead of closer to 40%.