AMRAAM G load

[sprstdlyscottsmn]

why is that CL go opposite to CN coefficient at a certain point ?

I can answer this part. CN is 90 degrees from the body, CL is 90 degrees from the airflow. So at 90deg AoA when CL is nearly 0 and CN is at max the CN is measuring drag only.

[eloise]

I can answer this part. CN is 90 degrees from the body, CL is 90 degrees from the airflow. So at 90deg AoA when CL is nearly 0 and CN is at max the CN is measuring drag only.

thanks you
so which one is a better indicator for turning ability ? CN or CL

[popcorn]

Trivia : AVEL subjects AMRAAM to 30G when ejected from F-22's belly bay.

[eloise]

I just found this ,

https://www.google.co.uk/url?sa=t&rct=j ... 1844,d.d24

they estimated the performance of Aim-120C-5 and estimate the CL value to be 0.11 at AoA of 30 degree and speed of 1200 m/s do you think that accurate ? ( sprut , John ? )

[johnwill]

I just found this ,

https://www.google.co.uk/url?sa=t&rct=j ... 1844,d.d24

they estimated the performance of Aim-120C-5 and estimate the CL value to be 0.11 at AoA of 30 degree and speed of 1200 m/s do you think that accurate ? ( sprut , John ? )

I don't see anything surprising about your CL vs CN chart. They are very similar up to 30 deg AoA, as I stated earlier. They have opposite signs at some angles since the reference axis for CL is flight path and for CN, it is missile body axis, which rotates with AoA.

The value of 0.11 for CL at 30 deg AoA cannot be compared to airplane lift coefficients since they are based on different reference areas. Airplanes normally use nominal wing area as the reference area for all aerodynamic coefficients, while the 0.11 missile CL is based on total wetted area of the missile (found on page 13 of the .pdf you posted (very interesting - thanks)). You may not understand wetted area - it is the total surface area of the missile and all its fins. The term "wetted" comes from boat technology, where they use the wet surface area of a boat as a coefficient reference area. Boat technology is of course much older than airplane technology, so sometimes airplane engineers borrow boat engineering terminology.

Note that the CFD analysis in your .pdf is limited to 30 deg AoA. Note from your CL and CN chart, both coefficients drop rapidly above 30 degrees, so maneuverability drops also. Drag is also much higher at higher AoA. From figure 14 in the .pdf, drag at 1000 m/s velocity is four times higher at 15 deg AoA than at zero AoA.

Since AoA is 30 deg max, either CL or CN may be used to precdict turn performance. Normally, as in the .pdf, CL is used.

[eloise]

I don't see anything surprising about your CL vs CN chart. They are very similar up to 30 deg AoA, as I stated earlier. They have opposite signs at some angles since the reference axis for CL is flight path and for CN, it is missile body axis, which rotates with AoA.

The value of 0.11 for CL at 30 deg AoA cannot be compared to airplane lift coefficients since they are based on different reference areas. Airplanes normally use nominal wing area as the reference area for all aerodynamic coefficients, while the 0.11 missile CL is based on total wetted area of the missile (found on page 13 of the .pdf you posted (very interesting - thanks)). You may not understand wetted area - it is the total surface area of the missile and all its fins. The term "wetted" comes from boat technology, where they use the wet surface area of a boat as a coefficient reference area. Boat technology is of course much older than airplane technology, so sometimes airplane engineers borrow boat engineering terminology.
.

Thanks alot John , really helpful

btw so the wetted area basically mean wing + body lift right ?
if i know the wetted area , CL coefficient , missiles weight , i can predict maximum instantaneous turn rate a missiles can make at certain speed-altitude using this formula : Lift = 1/2 * rho * v^2 * S * CL correct ?
For example :
Air density at 60K ft = 0.12 kg / m3
AIM-120 total weight is 152 kg , 52 kg of solid propellant so at burn out aim-120 weight will be 100 kg
at 60K feet speed of sound is 293 m/s
let say aim-120 max speed at burn out is mach 4
CL value at AoA of 30 degree = 0.11

at maximum instantaneous G turn then lift must be equal weight* acceleration so :
100*( G-load ) = 1/2* 0.12* (293*4 )^2 * (wetted area )* 0.11
am i doing it correct ? if the missiles motor still burning should i factor in missiles thrust as well ? given the fact that missiles T/W is alot higher than aircraft and at high AoA part of the thrust also useful as lift

[johnwill]

"wetted area" does not necessarily mean fin + body lift, as it is just a reference area used to calculate coefficients. In this case the CL does include lift from fin + body.

Your equations look OK to me, so whatever you calculate for "g" should be a good estimate. You should include the lift component of engine thrust as part of total lift. But what about the lift component of drag, shown in figure 14 of the .pdf? Should you include that? The answer is no, because figure 14 drag is in the flight path axis system, thus perpendicular to lift, with no lift component.

After you calculate "g", then calculate ITR from g and velocity. Remember also, that ITR equation is good only for constant altitude turns. For any pitch up or down, a component of earth gravity should be added to "g" from lift, although it will be relatively small in the missile case. That's why turn rates for airplanes are for constant altitude turns only.

[eloise]

"wetted area" does not necessarily mean fin + body lift, as it is just a reference area used to calculate coefficients. In this case the CL does include lift from fin + body.

Your equations look OK to me, so whatever you calculate for "g" should be a good estimate. You should include the lift component of engine thrust as part of total lift. But what about the lift component of drag, shown in figure 14 of the .pdf? Should you include that? The answer is no, because figure 14 drag is in the flight path axis system, thus perpendicular to lift, with no lift component.

After you calculate "g", then calculate ITR from g and velocity. Remember also, that ITR equation is good only for constant altitude turns. For any pitch up or down, a component of earth gravity should be added to "g" from lift, although it will be relatively small in the missile case. That's why turn rates for airplanes are for constant altitude turns only.

thanks you john your insight is very valuable
unfortunately though , the Pdf files didnt give any specific number for wetted area so i cant finished my equation






anyway according to the pdf file :

Ammendment: Source for the mass of the fuel has been found and now known to be 51.2559kg[7]
.
However, the below information is what was used for all the rest of the document.
To progress with the project, an assumption had to be made based off of all reasonable information.
Ultimately it was a fuel mass of 50kg, 265Isp and a 7.75s burn time providing the missile with a total
thrust of 16771.9N of thrust.

missiles put out total 16771N of thrust in 7.75 second so amount of thrust each second is 2164 Newtons ( i dont know where i made a mistake but this number seem very low for me , with thrust of only 2164 Newtons then the Aim-120 will have T/W of barely over 1.4 not that much better than fighter aircraft )
anyways
so lift equal thus*sin (AoA )
the lift from thrust when missiles fly at AoA of 5 degrees = 188N
the lift from thrust when missiles fly at AoA of 10 degrees = 375N
the lift from thrust when missiles fly at AoA of 15 degrees = 560N
the lift from thrust when missiles fly at AoA of 20 degrees = 704N
the lift from thrust when missiles fly at AoA of 30 degrees = 1082N

[johnwill]

Two comments--
1. You already calculated fin area earlier in this thread. Fin wetted area is two times normal area. It should be a simple matter to estimate body surface wetted area from length and diameter, making an adjustment for the nose cone surface area.

2. If thrust is 16771 N, why divide by 7.75 seconds? If you have impulse (thrust x time) you would divide by time to get thrust. But if you already have thrust, that's it.

[eloise]

Fin wetted area is two times normal area.

Can you explain this bit ? why wetted area of fin is 2 time normal area ?

It should be a simple matter to estimate body surface wetted area from length and diameter, making an adjustment for the nose cone surface area.

so i put picture of Aim-120 into paint and use the ruler app to calculate the ratio between the cone length and total missiles length
the pointy part of the cone accounted for 1/9.4 of total missiles length , so the cone length is 388 mm ( 0.388 meters ), missiles length is 3262 mm ( 3.262 meters )
missiles have diameter of 180 mm so the radius of the base is 90 mm ( 0.09 meters )
solved for area ( put number in formula )we have :
lateral surface area of Aim-120 body is : 1.84 square meters
lateral surface area of Aim-120 pointy nose cone is : 0.11 square meters
one question though when calculated wetted area do we only count the surface that directly hit the air flow ? ( for example : when missiles fly at a positive AoA then only the lower part of it's body hit the air flow while the upper part are not hit by anything )

2. If thrust is 16771 N, why divide by 7.75 seconds? If you have impulse (thrust x time) you would divide by time to get thrust. But if you already have thrust, that's it.

my bad so that why it seem so low , i assumed that they mean total amount of thrust in 7.75 second is 16771 N that why i divided it to get thrust in each second silly me
here the corrected value :

thrust is 16771 N
lift equal thus*sin (AoA )
the lift from thrust when missiles fly at AoA of 5 degrees = 1461 Newtons
the lift from thrust when missiles fly at AoA of 10 degrees = 2912 Newtons
the lift from thrust when missiles fly at AoA of 15 degrees = 4340 Newtons
the lift from thrust when missiles fly at AoA of 20 degrees = 5736 Newtons
the lift from thrust when missiles fly at AoA of 25 degrees = 7087 Newtons
the lift from thrust when missiles fly at AoA of 30 degrees = 8385.5 Newtons

[johnwill]

Wetted area of a fin is two times fin area because it gets wet on both sides when you put it in water. When we say fin area, we mean the projected area. When we say wetted area we mean the actual surface area of the total fin, both sides. The wetted area of a cube is six times the surface area of one side of the cube.

[eloise]

Wetted area of a fin is two times fin area because it gets wet on both sides when you put it in water. When we say fin area, we mean the projected area. When we say wetted area we mean the actual surface area of the total fin, both sides. The wetted area of a cube is six times the surface area of one side of the cube.

Thanks john
so here is my estimation :
I estimate the fin size of aim-120 earlier in this thread , however since I got the diameter and total fin span of the missiles wrong ,I used value of 178 mm for missiles diameter and wing span of 445 mm but according to the pdf file aim-120 diameter is 180 mm and fin span is 482 mm , I will do it again with same step but with a tiny change in number

AIM-120C-5 have body diameter of 180 mm and total wing span of 482 mm ( 0.482 meters ) which mean the span of each fin is ( 482-180)/2 = 151 mm ( 0.151 meters )
This is AIM-120C fin

here is the fin size relative to the total body


The photos shows that Aim-120's fin have trapezoid shape ,
For the rear fin : the big base length is around 2.5 times the height
the small base length is around 1.5 times the height
the height (wing span ) as we measured earlier is around 0.151 meters , so the big base is 0.3775 meters , small base is 0.2265 meters
Solve for area we have 0.046 square meters for each rear fin


For the frontal fin : the big base length is around 2 times the height
the small base length is around 0.5 times the height ,the height (wing span ) as we measured earlier is around 0.151 meters so the big base is 0.302 meters , small base is 0.0755 meters
Solve for area we have 0.029 square meters for each frontal fin

we have total of 4 frontal fins and 4 rear fins so total fin area on aim-120 is :
0.046*4 + 0.029*4 = 0.3 m2 ( this number is 3 times bigger than my estimation earlier in this thread because before I only estimate area of horizontal fin and ignore vertical fins )

as calculated earlier lateral surface area of Aim-120 body is : 1.84 square meters
lateral surface area of Aim-120 pointy nose cone is : 0.11 square meters

So the total wetted area of aim-120 is 1.84 + 0.11 + 0.3*2 = 2.55 square meters



According to the model graph in pdf file when altitude rise by 15 km the CL value at same AoA reduced by around 10% , Also from the graph : at AoA of 30 degree , altitude 5 km , velocity 1200 m/s then CL value = 0.11
From data above it safe to assumed at altitude of 60-65K feet ( 19-20 km ) , speed of 1200 m/s and AoA of 30 degree then CL value for wetted area of Aim-120 is around 0.099 - 0.1

[eloise]

put everything estimation value above into equation
Air density at 60K ft = 0.12 kg / m3
AIM-120 total weight is 152 kg , 52 kg of solid propellant so at burn out aim-120 weight will be 100 kg
at 60K feet speed of sound is 293 m/s
let say aim-120 max speed at burn out is mach 4
CL value at AoA of 30 degree , 60K feet = 0.1
total wetted area of aim-120 is 1.84 + 0.11 + 0.3*2 = 2.55 square meters

Lift = 1/2 * rho * v^2 * S * CL
at maximum instantaneous G turn then lift must be equal weight* acceleration so :
100*10*( G-load ) = 1/2* 0.12* (293*4 )^2 * (2.55)* 0.1
G-load = 21 G

I think I made some mistake some where again how the heck can a Aim-120 make an instantaneous turn of 21G at 60K feet at burn out ?

[garrya]

put everything estimation value above into equation
Air density at 60K ft = 0.12 kg / m3
AIM-120 total weight is 152 kg , 52 kg of solid propellant so at burn out aim-120 weight will be 100 kg
at 60K feet speed of sound is 293 m/s
let say aim-120 max speed at burn out is mach 4
CL value at AoA of 30 degree , 60K feet = 0.1
total wetted area of aim-120 is 1.84 + 0.11 + 0.3*2 = 2.55 square meters

Lift = 1/2 * rho * v^2 * S * CL
at maximum instantaneous G turn then lift must be equal weight* acceleration so :
100*10*( G-load ) = 1/2* 0.12* (293*4 )^2 * (2.55)* 0.1
G-load = 21 G

I think I made some mistake some where again how the heck can a Aim-120 make an instantaneous turn of 21G at 60K feet at burn out ?

This look great , can I bring this to Keypub ? it will solve our discussion related to how many G's the missile can pull at certain altitude and speed

[johnwill]

eloise,
Since I am more accustomed to working with English units (pounds, feet, etc), I converted everything to English units and got approximately the same answer. 21g seems reasonable to me, as I have heard of some missiles having a 50 g capability.