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 Subject: A simple formula for estimating missile performance. dwightlooi    3/15/2008 10:28:12 PM 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/sThis 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/sThe 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.

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 displacedjim       3/15/2008 11:26:58 PM And just to be clear, that's an estimate of the Vmax, or peak velocity.  In an all-boost motor of typical AAM size, this will be reached within about the first five seconds or so.  After that it's all deceleration as the the missile coasts for the rest of its fly-out and drag acts to slow it down.  For the longer ranged missiles with a boost-sustain motor, the peak velocity at burnout will come at most about 10 seconds later, or after about 10 to 15 seconds total burn time.  Again, after that it's all deceleration.  From the charts I've seen, I'd rough wag that basically all AAMs will be back down to under M3.0 within only about 10 to 15 seconds of coasting after burnout, and under M2.0 within about 20-30 seconds after that.

 dwightlooi       3/16/2008 2:00:01 PM And just to be clear, that's an estimate of the Vmax, or peak velocity.  In an all-boost motor of typical AAM size, this will be reached within about the first five seconds or so.  After that it's all deceleration as the the missile coasts for the rest of its fly-out and drag acts to slow it down.  For the longer ranged missiles with a boost-sustain motor, the peak velocity at burnout will come at most about 10 seconds later, or after about 10 to 15 seconds total burn time.  Again, after that it's all deceleration.  From the charts I've seen, I'd rough wag that basically all AAMs will be back down to under M3.0 within only about 10 to 15 seconds of coasting after burnout, and under M2.0 within about 20-30 seconds after that. 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.

 dwightlooi       3/16/2008 2:05:23 PM 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. Ooops! I meant to say "It'll take a missile a heck of a lot longer to decelerate from Mach 2 to Mach 1 compared to say Mach 4 to Mach 2 for instance."

 displacedjim       3/16/2008 4:24:48 PM And just to be clear, that's an estimate of the Vmax, or peak velocity.  In an all-boost motor of typical AAM size, this will be reached within about the first five seconds or so.  After that it's all deceleration as the the missile coasts for the rest of its fly-out and drag acts to slow it down.  For the longer ranged missiles with a boost-sustain motor, the peak velocity at burnout will come at most about 10 seconds later, or after about 10 to 15 seconds total burn time.  Again, after that it's all deceleration.  From the charts I've seen, I'd rough wag that basically all AAMs will be back down to under M3.0 within only about 10 to 15 seconds of coasting after burnout, and under M2.0 within about 20-30 seconds after that. 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.   I said "20-30 seconds after that."  The "that" I was trying to refer to was the attaining of M3.0.  From the curves I've seen (and from memory), I would roughly estimate after dropping to M3.0 it will typically take another 20-30 seconds to drop to M2.0.   I know the last time you and I talked about burn times you mentioned something about some sort of rocket motor/propellent design that gives a long burn, possibly even controllable duration burns. 1.  I don't claim to be familiar with state-of-the-art western rocket motor tech, so if that's the sort of motor that Raytheon, for example, is looking at putting into some new missile designs, then I wouldn't know about that. 2.  All the threat AAMs in the world today and all the Blue AAMs I'm familiar with (which I think covers what's currently in use) are exactly like I said.

 dwightlooi       3/16/2008 7:27:02 PM 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 AP = Density of Air (kg/m^3) ; ~1.29 kg/m^3 @ sea level; ~0.232 kg/m^3 @ 12,000 mV = Velocity (m/s) ; Mach 1 = 340 m/s @ sea level; ~295 m/s @ 12,000 mCd = Co-efficient of Drag ; ~ 0.6 to 0.95 for rockets depending mostly on finnage, nose and tail profileA = 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 lbsDrag Force @ Mach 2 = 0.5 x 0.232 x (295x2)^2 x 0.70 x 0.025 = 707 Newtons = 159 lbsDrag Force @ Mach 1 = 0.5 x 0.232 x 295^2 x 0.70 x 0.025 = 177 Newtons = 39.8 lbsThe 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 lbsAssuming 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 ftDeceleration @ 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 ftDeceleration @ Mach 3 = -F / mass = -11744 / (203 x 0.454) = -127 m/s^2 = - Mach 0.39/sec @ sea levelDeceleration @ 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 levelOK... 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 t

 displacedjim       3/16/2008 8:01:01 PM Very kewl to see how the numbers relate to each other.  Great point about the altitude, the reports I see are all standardized to specific sets of conditions, one of which is altiutudes of the launch and target aircraft.  I should have said something about that.  We can see from the equation and your examples how future missiles can be designed to really stretch their legs, and use in a high-altitude engagement really adds to any missile's range by a huge factor.  That's why the launch aircraft's altitude is a very important factor in determining the range of a shot, and one of the many important benefits for the F-22 from being able to routinely operate at 40,000ft, and supercruise even higher.  Thanks for sharing this info, Dwightlooi.

 EW3       3/16/2008 8:01:06 PM Where is the effect of gravity in the equations?  The missile is going down from the moment it is launched due to gravity unless you have thrust vectoring or lift.

 dwightlooi       3/16/2008 11:16:31 PM Where is the effect of gravity in the equations?  The missile is going down from the moment it is launched due to gravity unless you have thrust vectoring or lift. (1) In the absence of lift, the missile will accelerate downwards until it reaches a terminal vertical velocity. But that doesn't change the lateral velocity.(2) I didn't want to complicate the discussion. But let it suffice to say that At Mach 3, even with a total absence of wings, the amount of AoA the body needs to present to generate sufficient lift to keep the missile at a constant altitude is miniscue -- try a fraction of a degree. The amount of drag is from lift generation is also miniscue in the scope of the weapon being boosted at a 10:1 power to weight ratio or the aerodynamic drag forces at high Mach numbers. For the most parts, wings are not worth the drag they produce once a missile is sufficiently fast (say Mach 2+), this is why some designs like the ASRAAM which emphasizes kinematics do away with it and rely solely on body lift. The are useful however for enabling a rapid change in direction immediately post launch so less boost time is wasted on acceleration towards a wrong direction. This is why missiles which emphasizes off boresight engagements (WVR dogfight AAMs) tend to have big finnage or thrust vectoring, or both.(3) Naturally, gravity comes to play in a missile's performance, when shooting from 40,000 ft down to 20,000 ft the missile will add a fraction of gravitational acceleration to its own. When shooting up from 20,000 feet to 40,000 ft gravity robs a missile of its performance.

 dwightlooi       3/19/2008 7:58:23 AM 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 LevelRemember, 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%.http://www.aerospaceweb.org/question/atmosphere/density/density.jpg">