What is the Terminal velocity of a Ballistic Missile Warhead? Mach 17 here.

[Safriz]

The question of how fast a Ballistic missile's warhead is after reentering the atmosphere is not simple to answer as too many variables are involved.
But on average a Russian Yars suffers 30% deceleration and American Minuteman-3 suffers 44% deceleration on average after reentry. So we can take a grand average of 37% deceleration .Also it is estimated that an Indian ABM will have maximum of 90 seconds to hit a Shaheen-2 or Shaheen-3 / Ababeel warhead.Explanation below.
To make matters simple i have done the following simple calculations. Here in the video we see Russian MIRV-falling on targets . They are entering atmosphere at an angle ans i have assumed the angle to be 45 degrees although seems more.
So by assuming the angle of entry as 45 degrees from horizontal or vertical plane,we can do a simple right angle triangle calculation.
(Russian RS-24 Yars video)

The Warheads start glowing upon atmospheric entry which happens to be at 100 Km from ground. So the vertical side of the right angle triangle is taken as 90 Km because 10 Km of atmospheric friction allowed to make the warheads start glowing.
From the first light , which also happens to be the start of the video, the warheads seem to hit the ground in 21 seconds, or the first warhead falls in 21 seconds,the others fall at different times until 25 seconds.
So if we calculate the hypotenuse of the right angle triangle with perpendicular side of 90 Km and top bottom angles of 45 degrees, we will know the distance traveled by warheads after entering atmosphere and until hitting the ground? and keeping in mind the total time taken is 21 seconds we can calculate speed?
So from my calculations the first warhead travels 127 Km inside atmosphere in 21 seconds.
That's an average speed of 6 Kilometers per second. Thats a speed of about mach 17.
This was a Russian ICBM which enters atmosphere at about mach 20-22. So it has slowed down 3-5v Mach numbers as a result of atmospheric friction or a 30% loss of speed after reentry.
Just some simple logical calculations/

Here is another detailed video of USA's Minuteman -III. Watch at 0:54, 5:20, 6:01 seconds, it shows re-entry vehicles prior to impact. Narrator says speed is 16000 Km/h or Mach 13 after Reentry and before Impact . Which fairly matches with the calculations i did above.

Implications for Pakistan.

The Re-Entry speed of a Shaheen-2 Missile is about Mach 13 and that of a Shaheen-3 Missile is Mach 17-18.
By applying above assumptions and subtracting 37% of speed , the shaheen-2 will travel inside atmosphere at average speed of Mach 4.8. A Shaheen-3 and Ababeel (Using same stages and warheads) will travel inside atmosphere at average speed of Mach 6.
Time spent inside atmosphere will be about 70-90 seconds before impact. (Because Yars warheads spend 21-25 seconds inside atmosphere before Impact and are 3 times faster than Shaheen-3 assumed speed after reentry).
So any Indian ABM will have 90 seconds to hit an incoming Shaheen-3 or Shaheen-2 Warheads.

For further references here an Indian Agni-1 RV or Warhead can be seen after Reentry. You can guess the speed? (Watch after 1:25) The flash you see is the detonation of trigger mechanism, the only such video publicly available in which a missile's nuclear warhead's trigger is being shown activating.

 

[Safriz]

@amardeep mishra @HRK @The Deterrent and others interested

 

[Little Falcon]

Well according to my knowledge, terminal velocity depends upon mass of an object, its cross sectional area and how smooth it can pass through fluid.

You can yourself calculate terminal velocity by using formula: v = the square root of ((2*m*g)/(ρ*A*C)).

• m = mass of the falling object
• g = the acceleration due to gravity. On Earth this is approximately 9.8 meters per second per second.
• ρ = the density of the fluid the object is falling through.
• A = the projected area of the object. This means the area of the object if you projected it onto a plane that was perpendicular to the direction the object is moving.
• C = the drag coefficient. This number depends on the shape of the object. The more streamlined the shape, the lower the coefficient. You can look up some approximate drag coefficients here.

 

[Safriz]

Well according to my knowledge, terminal velocity depends upon mass of an object, its cross sectional area and how smooth it can pass through fluid.

You can yourself calculate terminal velocity by using formula: v = the square root of ((2*m*g)/(ρ*A*C)).

• 
  
  
  • m = mass of the falling object
  • g = the acceleration due to gravity. On Earth this is approximately 9.8 meters per second per second.
  • ρ = the density of the fluid the object is falling through.
  • A = the projected area of the object. This means the area of the object if you projected it onto a plane that was perpendicular to the direction the object is moving.
  • C = the drag coefficient. This number depends on the shape of the object. The more streamlined the shape, the lower the coefficient. You can look up some approximate drag coefficients here.

Well i cannot use this formula as too many variables are unknown (and lame at math). Thats why invented a simplistic approach.
But if i say the Reentry speed of a Shaheen-3 warhead is a cone which enters atmosphere at Mach 17. Height is 2 meters Base Diameter is 0.75 meters. Mass is 700 Kg or less , can you run some calculations?

 

[CriticalThought]

Well according to my knowledge, terminal velocity depends upon mass of an object, its cross sectional area and how smooth it can pass through fluid.

You can yourself calculate terminal velocity by using formula: v = the square root of ((2*m*g)/(ρ*A*C)).

• 
  
  
  • m = mass of the falling object
  • g = the acceleration due to gravity. On Earth this is approximately 9.8 meters per second per second.
  • ρ = the density of the fluid the object is falling through.
  • A = the projected area of the object. This means the area of the object if you projected it onto a plane that was perpendicular to the direction the object is moving.
  • C = the drag coefficient. This number depends on the shape of the object. The more streamlined the shape, the lower the coefficient. You can look up some approximate drag coefficients here.

g, rho, and C will keep changing over time as the warhead gets nearer to earth. m could change if the outer layers peel off due to friction. A could change if the warhead gets deformed because of all the forces acting on it.

@شاھین میزایل I once read in a book on signals and systems that a rocket's trajectory is governed by 16th or 17th order differential equations. Simplistic calculations won't provide any degree of accuracy for the performance of a warhead's re-entry.

 

[amardeep mishra]

I've answer it in one of the threads.refer to my simulations

 

[amardeep mishra]

The question of how fast a Ballistic missile's warhead is after reentering the atmosphere is not simple to answer as too many variables are involved.
But on average a Russian Yars suffers 30% deceleration and American Minuteman-3 suffers 44% deceleration on average after reentry. So we can take a grand average of 37% deceleration .Also it is estimated that an Indian ABM will have maximum of 90 seconds to hit a Shaheen-2 or Shaheen-3 / Ababeel warhead.Explanation below.
To make matters simple i have done the following simple calculations. Here in the video we see Russian MIRV-falling on targets . They are entering atmosphere at an angle ans i have assumed the angle to be 45 degrees although seems more.
So by assuming the angle of entry as 45 degrees from horizontal or vertical plane,we can do a simple right angle triangle calculation.
(Russian RS-24 Yars video)

The Warheads start glowing upon atmospheric entry which happens to be at 100 Km from ground. So the vertical side of the right angle triangle is taken as 90 Km because 10 Km of atmospheric friction allowed to make the warheads start glowing.
From the first light , which also happens to be the start of the video, the warheads seem to hit the ground in 21 seconds, or the first warhead falls in 21 seconds,the others fall at different times until 25 seconds.
So if we calculate the hypotenuse of the right angle triangle with perpendicular side of 90 Km and top bottom angles of 45 degrees, we will know the distance traveled by warheads after entering atmosphere and until hitting the ground? and keeping in mind the total time taken is 21 seconds we can calculate speed?
So from my calculations the first warhead travels 127 Km inside atmosphere in 21 seconds.
That's an average speed of 6 Kilometers per second. Thats a speed of about mach 17.
This was a Russian ICBM which enters atmosphere at about mach 20-22. So it has slowed down 3-5v Mach numbers as a result of atmospheric friction or a 30% loss of speed after reentry.
Just some simple logical calculations/

Here is another detailed video of USA's Minuteman -III. Watch at 0:54, 5:20, 6:01 seconds, it shows re-entry vehicles prior to impact. Narrator says speed is 16000 Km/h or Mach 13 after Reentry and before Impact . Which fairly matches with the calculations i did above.

Implications for Pakistan.

The Re-Entry speed of a Shaheen-2 Missile is about Mach 13 and that of a Shaheen-3 Missile is Mach 17-18.
By applying above assumptions and subtracting 37% of speed , the shaheen-2 will travel inside atmosphere at average speed of Mach 4.8. A Shaheen-3 and Ababeel (Using same stages and warheads) will travel inside atmosphere at average speed of Mach 6.
Time spent inside atmosphere will be about 70-90 seconds before impact. (Because Yars warheads spend 21-25 seconds inside atmosphere before Impact and are 3 times faster than Shaheen-3 assumed speed after reentry).
So any Indian ABM will have 90 seconds to hit an incoming Shaheen-3 or Shaheen-2 Warheads.

For further references here an Indian Agni-1 RV or Warhead can be seen after Reentry. You can guess the speed? (Watch after 1:25) The flash you see is the detonation of trigger mechanism, the only such video publicly available in which a missile's nuclear warhead's trigger is being shown activating.

Hi @Shaheen missile!
The Re-entry isnt all that complicated if you knew aerospace and how to solve non-linear and coupled differential equations. Contrary to what you might assume,a RV kinematics eqns are of 4th order that can be numerically solved using any solver utilizing RK-4 or RK-5 like odeint in python and ode45 in MATLAB. The accuracy of such a simulation drastically depends on the atmospheric model you choose. Following are 4 differential eqns that one must solve simultaneously to get the trajectory information-

Now the single most important parameter that determines your RV kinematics is "beta" or ballistic coefficient.It is also important to pay attention to the fact that RV of missiles follow what is known as "ballistic trajectory" instead of "lifting trajectory".I wont delve too much into the details of each and suggest you to look the difference between the two in any decent aerospace book. Lets pay little bit more attention to "beta".
beta=W/Cd*A.
Which means for a higher beta we would ideally want as low Drag and effective Areas as possible(keeping the weight constant).A higher beta improves terminal velocity which in turns improves the accuracy of the RV as your RV has lesser tendency to get affected by cross winds etc.
Some of the important point that can be inferred by solve the above system of 4 de-eqns-
1)A lower beta results in less intense heating and deceleration whereas a higher beta yields correspondingly higher heating and deceleration.

Modern american ICBMs have beta in the vicinity of 24,500kg/m^2 and they yield something like 7-8Mach upon impact.I highly doubt any Indian or Pakistani RV have beta even close to this figure(I firmly believe that indian strategic missiles especially the latest ones are a generation ahead of their Pakistani counterparts).A more probable figure for pakistani RVs would be something like 8000-10000kg/m^2 which would yield something like 3machs on impact.
I'd also like to point out that flight path angle at re entry is kept shallow--more like 10-12degrees and not 45 degrees as your assuming. A smaller flight path angle results in graceful generation of heat instead of steep flight path angle.
Kindly note: flight path angle is the angle between wind axis system and your inertial system and not between body axis and inertial.

 

[amardeep mishra]

once read in a book on signals and systems that a rocket's trajectory is governed by 16th or 17th order differential equations.

Sorry but books on signals and systems do not deal with differential equations describing kinematics of RV.for that you'd better need a good book on orbital mechanics or re entry dynamics.

 

[Little Falcon]

Well i cannot use this formula as too many variables are unknown (and lame at math). Thats why invented a simplistic approach.
But if i say the Reentry speed of a Shaheen-3 warhead is a cone which enters atmosphere at Mach 17. Height is 2 meters Base Diameter is 0.75 meters. Mass is 700 Kg or less , can you run some calculations?

Based on your question, its simply inaccurate to use this formula as mentioned by @CriticalThought . As the constants are actually variable. So in-order to solve your question you have to make use of a simulation software. In case if you want to learn more check this: link

 

[amardeep mishra]

@The Deterrent
Here is what I'm doing today--

 

[Safriz]

Hi @Shaheen missile!
The Re-entry isnt all that complicated if you knew aerospace and how to solve non-linear and coupled differential equations. Contrary to what you might assume,a RV kinematics eqns are of 4th order that can be numerically solved using any solver utilizing RK-4 or RK-5 like odeint in python and ode45 in MATLAB. The accuracy of such a simulation drastically depends on the atmospheric model you choose. Following are 4 differential eqns that one must solve simultaneously to get the trajectory information-
View attachment 384680 View attachment 384681
Now the single most important parameter that determines your RV kinematics is "beta" or ballistic coefficient.It is also important to pay attention to the fact that RV of missiles follow what is known as "ballistic trajectory" instead of "lifting trajectory".I wont delve too much into the details of each and suggest you to look the difference between the two in any decent aerospace book. Lets pay little bit more attention to "beta".
beta=W/Cd*A.
Which means for a higher beta we would ideally want as low Drag and effective Areas as possible(keeping the weight constant).A higher beta improves terminal velocity which in turns improves the accuracy of the RV as your RV has lesser tendency to get affected by cross winds etc.
Some of the important point that can be inferred by solve the above system of 4 de-eqns-
1)A lower beta results in less intense heating and deceleration whereas a higher beta yields correspondingly higher heating and deceleration.View attachment 384705
Modern american ICBMs have beta in the vicinity of 24,500kg/m^2 and they yield something like 7-8Mach upon impact.I highly doubt any Indian or Pakistani RV have beta even close to this figure(I firmly believe that indian strategic missiles especially the latest ones are a generation ahead of their Pakistani counterparts).A more probable figure for pakistani RVs would be something like 8000-10000kg/m^2 which would yield something like 3machs on impact.
I'd also like to point out that flight path angle at re entry is kept shallow--more like 10-12degrees and not 45 degrees as your assuming. A smaller flight path angle results in graceful generation of heat instead of steep flight path angle.
Kindly note: flight path angle is the angle between wind axis system and your inertial system and not between body axis and inertial.

Hi,
Watch the second video. The narrator says the speed of rv as 16000 kph which is much higher than Mach 8?

 

[amardeep mishra]

Hi,
Watch the second video. The narrator says the speed of rv as 16000 kph which is much higher than Mach 8?

Hi shaheen~
The narrator hasnt probably done any simulations.In case you need proper research papers,do let me know,I will fill you up with plethora of research papers on RV kinematics and dynamics. Instead of going by such "narrations" why dont you try proper research papers from AIAA or IEEE for a change?Believe me you will end up learning much more.ALthough for such an en devour you need pretty decent mathematics and perhaps coding abilities.
If you look at the graphs above,the RV velocity at height of around 70-80kms is close to 22-23Machs.The velocity just before entering into atmosphere is again dependent on the velocity of the RV at "burn-out".More the velocity at burnout,more will be the velocity just before re-entry into atmosphere.However as the RV descends into atmosphere,it looses kinetic energy and slows down.By the time RV actually reaches 0-500m altitude,it has come down to almost 7M(in the case of modern american ICBMs) and ~3M(in the case of pakistani IRBMs). So,below 15kms altitude,pakistani IRBMs would have most likely come down to 7-9M as can be seen from the graphs above.
The main point here is,interception doesnt take place as you're thinking,I will try to briefly describe it as a Control Engineer!
1)The main surveillance radar(DRDO's swordfish based on ELTA's greenpine) scans the volume using fan beams.Upon detection of any RV,the target information is
2)Passed onto a Fire control radar(MFTR) which can track a fast moving RV much more precisely(with higher resolution) vis-a-vis detection radar. This radar tracks the RV and computes firing solution.
3)Firing solution corresponds to finding probable cordinates where the RV will be in some future time.This is slightly tricky.But this can be done "assuming" a projectile like trajectory of RV--after all an RV is projectile in atmosphere(i.e it doesnt manuvre inside atmosphere).
4)Once this has been computed,interceptors are fired at this "probable cordinate",the GNC module of missile takes over and it steers the missile to this coordinate. I wont go deeper into the GNC algorithms.But In case you want,I can write another thread.
5)As the interceptor reaches this coordinate,the IIR or active radar goes active and starts searching for the RV.After this,its all about steering the interceptor so that it hits the incoming RV(which is coming in an projectile like trajectory).

 

[Safriz]

Hi shaheen~
The narrator hasnt probably done any simulations.In case you need proper research papers,do let me know,I will fill you up with plethora of research papers on RV kinematics and dynamics. Instead of going by such "narrations" why dont you try proper research papers from AIAA or IEEE for a change?Believe me you will end up learning much more.ALthough for such an en devour you need pretty decent mathematics and perhaps coding abilities.
If you look at the graphs above,the RV velocity at height of around 70-80kms is close to 22M

Simpler thing will be if you can carry out simplistic calculations, write the values of variables in the formula so that we too can read and come up with a final value?

 

[amardeep mishra]

Simpler thing will be if you can carry out simplistic calculations, write the values of variables in the formula so that we too can read and come up with a final value?

I have edited my post above,kindly have a look again,and for "simplistic calculations" ,I am afraid ,you would have to solve differential equations to know the trajectory information.And for that you would need to know beta precisely,and for that you would need to know the weight,drag polar and effective area.
So in case you can provide these details,I can run the simulations to find the probable trajectory of shaheens.Anyhow I already ran a simulation long time back assuming a beta value of 8000kg/m^2 for pakistani RV and I got something like 3M at the time of impact.However to find the velocity of RV just before ren-entry you need to solve other equations.

 

[Safriz]

I have edited my post above,kindly have a look again,and for "simplistic calculations" ,I am afraid ,you would have to solve differential equations to know the trajectory information.And for that you would need to know beta precisely,and for that you would need to know the weight,drag polar and effective area.
So in case you can provide these details,I can run the simulations to find the probable trajectory of shaheens.Anyhow I already ran a simulation long time back assuming a beta value of 8000kg/m^2 for pakistani RV and I got something like 3M at the time of impact.However to find the velocity of RV just before ren-entry you need to solve other equations.

https://www.rand.org/content/dam/rand/pubs/research_memoranda/2008/RM3475.pdf

The Ballistic coefficient is way different in this research paper ....
Far lesser than the value you are mentioning?