How to calcualte Power Necessary to mantain speed while firing with axial weapon!
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How to calcualte Power Necessary to mantain speed while firing with axial weapon!
Hi every body.
I' m an helo pilot, and i'm triyng to calculate the Power necessary in terms of Torque to mantain speed while firing with axial weapon.
Let's suppose i'm cruising 90 Kts, with 85% of torque. I know the Helo mass; the calliber is 12.7 mm, 2 axial weapon. I know the mass and the exit speed of the bullet. How much torque do i need for a single fired bullet to counteract the back force, in order to mantain same speed?
Does anybody things would be possibile to calculate it, or is just simply fantasy?
Thank you in advance to everybody, that will provide useful infos.
I' m an helo pilot, and i'm triyng to calculate the Power necessary in terms of Torque to mantain speed while firing with axial weapon.
Let's suppose i'm cruising 90 Kts, with 85% of torque. I know the Helo mass; the calliber is 12.7 mm, 2 axial weapon. I know the mass and the exit speed of the bullet. How much torque do i need for a single fired bullet to counteract the back force, in order to mantain same speed?
Does anybody things would be possibile to calculate it, or is just simply fantasy?
Thank you in advance to everybody, that will provide useful infos.
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I think I would apply conservation of momentum...
If the plane was stationary when it fired...
MB x VB = MP x VP
MB = mass of bullet
VB = velocity of bullet
MP = mass of plane
VP = Recoil velocity of plane
but the plane isn't stationary so...
The airspeed after firing will be the airspeed before firing minus VP.
I think!
If the plane was stationary when it fired...
MB x VB = MP x VP
MB = mass of bullet
VB = velocity of bullet
MP = mass of plane
VP = Recoil velocity of plane
but the plane isn't stationary so...
The airspeed after firing will be the airspeed before firing minus VP.
I think!
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Do a Hover - it avoids G
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cwatters
That would be my approach too. Once you have the recoil force and the time it is applied (how long are you firing how many bullets etc) you can then apply this force and time to the mass of the chopper when stationary (f=ma) and get the acceleration that would result on the chopper. This would be a deceleration when in flight - but we know that.
Once you know that the chopper will loose say x kt in y secs (from the deceleration established above) you then check the chopper for the torque needed to gain the same number of kt above the chosen firing speed over the same length of time you intend to fire. As you say all simple Mr Newton really.
The result should be slightly pessimistic since the extra v squared drag of checking torque to increase speed above the firing speed would not be there when maintaing speed. I suppose once you know the number of kots lost firing at constant torque you could bracket firing speed to reduce this effect.
That would be my approach too. Once you have the recoil force and the time it is applied (how long are you firing how many bullets etc) you can then apply this force and time to the mass of the chopper when stationary (f=ma) and get the acceleration that would result on the chopper. This would be a deceleration when in flight - but we know that.
Once you know that the chopper will loose say x kt in y secs (from the deceleration established above) you then check the chopper for the torque needed to gain the same number of kt above the chosen firing speed over the same length of time you intend to fire. As you say all simple Mr Newton really.
The result should be slightly pessimistic since the extra v squared drag of checking torque to increase speed above the firing speed would not be there when maintaing speed. I suppose once you know the number of kots lost firing at constant torque you could bracket firing speed to reduce this effect.
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VP is the loss of velocity due to one round. If you know the number of rounds per second fired you can work out the deceleration in kts/second while firing.
The power required to maintain speed would be the power required to accelerate the plane at the same rate from V-VP back upto V.
Since we are trying to maintain airspeed we could assume the variation in drag force is small (eg drag is approx constant). All that has to be overcome to accelerate the plane is the mass of the plane.
Edit: This post crossed with Johns.
The power required to maintain speed would be the power required to accelerate the plane at the same rate from V-VP back upto V.
Since we are trying to maintain airspeed we could assume the variation in drag force is small (eg drag is approx constant). All that has to be overcome to accelerate the plane is the mass of the plane.
Edit: This post crossed with Johns.
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Not yet there!
Thank you cwatters for your replay.
I didn't know about the web page... but was 1st thing done. The problem, by the way is not yet solved. I still miss "how much power in terms of torque do i need to mantain initial speed"?
Thank you, and if you have other suggestion don't esitate.
BY
I didn't know about the web page... but was 1st thing done. The problem, by the way is not yet solved. I still miss "how much power in terms of torque do i need to mantain initial speed"?
Thank you, and if you have other suggestion don't esitate.
BY
Do a Hover - it avoids G
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Alex
That depends on the characteristics of your chopper about which we know nothing. Presumably you have all that information.
I would hazard a guess that unless you had a very heavy and powerful chopper you are bound to lose a knot or two.
That depends on the characteristics of your chopper about which we know nothing. Presumably you have all that information.
I would hazard a guess that unless you had a very heavy and powerful chopper you are bound to lose a knot or two.
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Alexpilot,
Not easy to work this out directly based on the information we normally have available for a helicopter. You can estimate this by assuming a small increase in torque (and hence a small increase in power) goes directly towards countering the firing effect. In truth, that's not how it will work because you'll also have to increase anti-torque, and potentially change the attitude of the helicopter.
If you're talking about high momentum rounds or high rates of fire, that can alter the attitude of the helicopter, then you'll also have to consider the power effects of changing the controls to counter this.
Once you work out the power required to maintain helicopter speed, you just use torque x rotor rpm to get power. Conversion will take some knowledge, but if you can set up a known power output from the engines, compare it to the torque produced, you'll get close. Of course, there are losses in the system, so if you need 1hp to counter the firing, you might actually need to demand 1.3hp more from your helicopter.
If you multiply the change in momentum caused by the gun by the velocity of the aircraft, you should get the power required to counter the firing effect.
Once all the calculations are complete, I'd guess you'd be underestimating torque by 20-50% because of the losses and because not all the torque is applied to counter the firing effect, but it will give you an idea.
I like John's method, but you might be able to go a step further. Without firing the weapon, measure the acceleration from the firing speed, due to a small increase in torque. Measure it for a number of small increases, and then you can establish the relationship between change in torque to acceleration rate. Match it to the predicted deceleration due to firing, and you should get a fairly close number for the require torque increase.
Excellent thread, there's a chance I'll be doing some of this testing in the not too distant future.
Matthew.
Not easy to work this out directly based on the information we normally have available for a helicopter. You can estimate this by assuming a small increase in torque (and hence a small increase in power) goes directly towards countering the firing effect. In truth, that's not how it will work because you'll also have to increase anti-torque, and potentially change the attitude of the helicopter.
If you're talking about high momentum rounds or high rates of fire, that can alter the attitude of the helicopter, then you'll also have to consider the power effects of changing the controls to counter this.
Once you work out the power required to maintain helicopter speed, you just use torque x rotor rpm to get power. Conversion will take some knowledge, but if you can set up a known power output from the engines, compare it to the torque produced, you'll get close. Of course, there are losses in the system, so if you need 1hp to counter the firing, you might actually need to demand 1.3hp more from your helicopter.
If you multiply the change in momentum caused by the gun by the velocity of the aircraft, you should get the power required to counter the firing effect.
Once all the calculations are complete, I'd guess you'd be underestimating torque by 20-50% because of the losses and because not all the torque is applied to counter the firing effect, but it will give you an idea.
I like John's method, but you might be able to go a step further. Without firing the weapon, measure the acceleration from the firing speed, due to a small increase in torque. Measure it for a number of small increases, and then you can establish the relationship between change in torque to acceleration rate. Match it to the predicted deceleration due to firing, and you should get a fairly close number for the require torque increase.
Excellent thread, there's a chance I'll be doing some of this testing in the not too distant future.
Matthew.
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Can it work?
I just finished to make calculation...!
Bottom line is the helo at determinated condition of weight, will loose approximatly 0,5 Km/h for each round fired. Lets suppose i fire 100 rounds, in "x" seconds i will lose 50 Km/h about 28 Kts... to make it easier... and worst lets say 30 Kts.
Now to calculate Torque, if we consider the last part (at relatively high speed 90Kts up to Vne) of the curve Power Required vs Velocity as straight line,so direct proportional (i know it will be quite a great approximation) if i read the Torque at 90 Kts, then accelerate up 120 Kts, read back the Torque, the difference between the 2 value could be the Torque necessary to mantain speed.
What do you think? Could be pretty resonable?
Thank you
Bottom line is the helo at determinated condition of weight, will loose approximatly 0,5 Km/h for each round fired. Lets suppose i fire 100 rounds, in "x" seconds i will lose 50 Km/h about 28 Kts... to make it easier... and worst lets say 30 Kts.
Now to calculate Torque, if we consider the last part (at relatively high speed 90Kts up to Vne) of the curve Power Required vs Velocity as straight line,so direct proportional (i know it will be quite a great approximation) if i read the Torque at 90 Kts, then accelerate up 120 Kts, read back the Torque, the difference between the 2 value could be the Torque necessary to mantain speed.
What do you think? Could be pretty resonable?
Thank you
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There is a very simple method to work this out by considering conservation of momentum over time.
Work out total forward momentum of helicopter (Mass x velocity) integrated over a Dt ( any standard time increment).
Work out total forward momentum of rounds integrated over the same Dt.
The second divided by the first will give you percentage that you have to increase torque to maintain same airspeed - assuming the discharge of rounds does not significantly affect the overall mass of the aircraft!
Work out total forward momentum of helicopter (Mass x velocity) integrated over a Dt ( any standard time increment).
Work out total forward momentum of rounds integrated over the same Dt.
The second divided by the first will give you percentage that you have to increase torque to maintain same airspeed - assuming the discharge of rounds does not significantly affect the overall mass of the aircraft!
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Maybe in the clear light of morning I could explain my above solution a bit better!
If gun fires n rounds per second, each mass m, at muzzle velocity v relative to helicopter, and the helicopter is mass M, flies at V, with torque C, then the increase in C necessary to maintain V will be nmvC/MV.
This assumes that M is constant - which is not strictly the case if n and m are very high...
If gun fires n rounds per second, each mass m, at muzzle velocity v relative to helicopter, and the helicopter is mass M, flies at V, with torque C, then the increase in C necessary to maintain V will be nmvC/MV.
This assumes that M is constant - which is not strictly the case if n and m are very high...
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If you wish to speak from what is the best 'user' aspect i.e. a purely practical point of view, then the answer self-evidently is that you will need a small increase in torque. If there is even a minor Delta IAS (CAS/EAS or whatever but for simplicity IAS) decrease per round fired then at N rounds fired IAS approaches zero and the power/ torque required to hover will be in excess of the power that is required when the helicopter is in unaccelerated flight at an IAS somewhere around the middle of the 'power required' specific altitude inverted bell curve; . RichA I think has got it right but remember; the mathematical constant in the kinetic energy equation is there because the theoretical equation calls for a measured difference between starting 'speed' and ending 'speed'. the 'half' is to approximate this difficult to measure number. And we have not as yet introduced the power effects of 'flap-back' and 'flap-forward' on the rotor disk. How is the 'torque' presented to you in the cockpit? And of course, we are ignoring the effects of decreasing IAS on the tail-rotor drag. Answer: Pull one or two more 'inches' (Western helicopter obviously) and see if the IAS increases during firing. Conversely, maintain MAP and see if the IAS decreases during firing but mantain a constant flight path to the offset aim point. Empirical?
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remember; the mathematical constant in the kinetic energy equation is there because the theoretical equation calls for a measured difference between starting 'speed' and ending 'speed'. the 'half' is to approximate this difficult to measure number.
First I've heard of that. Do you have a reference? This might have been useful when I was doing my post grad in particle physics.
Matthew.
First I've heard of that. Do you have a reference? This might have been useful when I was doing my post grad in particle physics.
Matthew.
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This must be the Junior School equation for kinetic energy; E = 1/2 x M x V squared. Where E is the Energy, M the Mass and V the Velocity at the instant of measurement. It is true that the constant of 0.5 does approximate very closely the value between V and static.
Last edited by GOLF_BRAVO_ZULU; 13th Dec 2007 at 12:49. Reason: Finger trouble
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A simultaneously-triggered, rearward-firing, gun with the same characteristics as the forward-pointing one will take the problem away... 
Other than for those behind you, of course!
Seriously, and notwithstanding the validity of your question, I suspect that the speed and agility requirements placed on a combat helicopter make this a very academic exercise.
From a practical point of view, flight test might be the quick and pragmatic approach, and should be without too many difficulties.
Other than for those behind you, of course!
Seriously, and notwithstanding the validity of your question, I suspect that the speed and agility requirements placed on a combat helicopter make this a very academic exercise.
From a practical point of view, flight test might be the quick and pragmatic approach, and should be without too many difficulties.
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It is true that the constant of 0.5 does approximate very closely the value between V and static
It is in fact exactly 0.5.
It's calculus in'it
pb
