Arrow passing through a moving target -- What is the final velocity?

[Warbow]

A 0.1 kg arrow is moving at 48 m/s to the right and strike a 3 kg object (helmet resting on a 5 kg head) moving to the left towards the arrow at 9 m/s (full mounted charge). It takes 80 Joules to penetrate the 3 kg helmet, what is the velocity of the arrow after penetration when it slams into the head? What is the velocity of the 3 kg helmet after penetration?

I'm stuck on this one. I know the KE of the arrow (115.2 Joule). I know the KE of the helmet (121.5 Joule). I know the momentum of the arrow (4.8 kg *m/s) and the object (27 kg * m/s).

I know how to calculate the final velocity of the helmet/arrow if the arrow get stuck in the helmet as an inelastic collision: (0.1*48+3*-9)/(0.1+3) = -7.1613 m/s

And from this that the KE of the object is now: 0.5*(0.1+3)*-7.1613^2 = 79.49 Joules.
Subtract this from (115.2+121.5) and you get 157.21 Joule going into penetration, heat and sound.

I have no idea when the arrow passes through the helmet.

How do I find out the amount of KE coming from the arrow vs helmet when dealing with those 80 Joules?

How does the 5 kg head have an impact on the outcome of this collision? I can't threat them as one solid object either. The helmet has a liner and as such reduces the impulse slightly.

I need a formula I can use in excel.

[Moderator's note: Moved from a technical forum and thus no template.]

 

[haruspex]

if the arrow get stuck in the helmet

Ah, but it doesn’t: it continues to pass through the helmet.
You are told how much energy goes into heat etc. What does that leave for KE of arrow+helmet?

It is not entirely clear whether the helmet is to be taken as loose fitting, as you have, or tight fitting. If the latter you need to include the head in the collision.

 

[kuruman]

I am trying to picture the situation from how you are trying to model it. It looks like you have an arrow that collides with a head wearing a helmet with a liner. Is the head catapulted towards the arrow? If so, what does "full mounted charge" mean?

 

[haruspex]

what does "full mounted charge" mean?

I think that is just to explain the 9m/s.

 

[Warbow]

This is not homework. I have no idea why it was moved to this forum. I worded that question. I am currently writing on a calculator in excel on arrow trajectory and air resistance using the Euler/trapezoid method. As of now it takes 3 seconds to do all the calculations. I will combine this with an armour penetration calculator which takes into account arrowhead shape, fracture toughness of the armour, angle of impact and armour thickness. Most of it is done. I have only a few things left. One of those problems has to do with what I asked about. I'm interested in the use of longbows in war from around 1270 to 1500 and beyond. I want to explain why they penetrated low quality plate armour of inadequate thickness (see the quotes at the end).

The problem is an arrow colliding with a helmet, penetrating through the helmet and colliding with the head. The helmet is on the head of a knight. The knight charges towards the arrow (and archer) on a horse at 9 m/s.
I want to know the kinetic energy going into the head. I want to know the speed of the arrow after penetration when it hits the head. I want to know the speed of the helmet moving in the opposite direction.

If the KE combined (helmet and arrow) is 236.7 Joule, then there is 156.7 Joule left?

If the target is stationary the helmet will move back at what speed? I think I can do that. 115.2 Joule - 80 Joule is 35.2 Joule left. SQRT((35.2*2)/0.1) = 26.53 m/s

(0,1*48+3*0)=(0.1*26.53)+3*v(helmet)
4.8=2.653+3*v(helmet)
4.8-2.653+3*v(helmet)
2.147=3*v(helmet)
2.147/3 = 3*v(helmet)/3
0.7156 = v(helmet)

Helmet is moving back at 0.7156 m/s.

In excel i will write it like this ((0.1*48+3*-0)-(0.1*26.53))/3

The lost energy is 0.5*3*0.7156^2 = 0.7682 Joule

115.2 - (80+0.7682) = 34.43 Joule left going into the target head.

Correct me if this is wrong.

So how do I find the speed of the arrow and helmet after the collision when the helmet is moving towards the arrow at 9 m/s and it takes 80 Joule to penetrate it.

To my defense I have never taken a physics class in my life.

The battle of Homildon Hill 1402.
The earl of Douglas, “evidently placing trust in his armor and that of his companions, which for three years they had taken pains to improve ... strove to rush the archers,” the bowmen “pierced entirely through these armored men [armatos omnino penetrarent], drilling through their helmets [cassides terebrarent] ... and piercing through all their armor with ease [et omnem armaturam levi negotio transverberarent]. The earl of Douglas was pierced [confossus est] with ve [5] wounds, notwithstanding his extremely costly [sumptuosissima] armor.” He lost an eye and a testicle in this battle.
Thomas Walsingham, Historia Anglicana ed. H. T. Riley, 2 vols. (London, 1864), 2:251–52.Thomas Walsingham, describing a battle of 1383, says that English archers ‘surpassed all others...for they so struck the enemy with their flying arrows that of their armored men no more remained [unharmed? on the field?] than if they had been unarmored... Bodies were perforated, their armor [lorica] notwithstanding; breasts were wounded, the plates [lamina] not resisting; heads were shot through [transfigebantur], the helmets not helping; hands holding lances or shafts were nailed to them, gauntlets being no protection.’
St Albans Chronicle, p. 680Here from the siege of Pontevedra in 1386.
The bailiff of the town was struck by an English arrow (qui luy percha le bacinet et la teste aussi) which pierced his bascinet and his head also.
Froissart, Oeuvres, 11:412.

The battle of Agincourt in 1415. "But the French nobility, who had previously advanced in line abreast and had all but come to grip with us, either from fear of the missiles which by their very force pierced the sides and visors of their helmets ..."
The Gesta Henrici Quinti (c. 1417, Latin), Chapter 13

 

[haruspex]

The helmet is on the head of a knight. The knight charges towards the arrow (and archer) on a horse at 9 m/s.
I want to know the kinetic energy going into the head.

Ok, but you have not answered my question about how snugly the helmet fits. I suggest it would be very snug, so you need to involve the mass of the head in the initial collision.

You have two known masses (arrow on the one hand, head+helmet on the other) and speeds before the collision, two unknown speeds for those two masses when the tip of the arrow reaches the head, and a presumed loss of KE. Using the two conservation laws you can deduce the two speeds.

It is very hard to follow working when numbers are plugged in at the start. Please define and use names for the constants, only plugging in numbers at the end. (This has many advantages.)

 

[Warbow]

haruspex

I was wondering about that too. At first I thought that I should just calculate it as a pendulum with a 3 kg helmet and a 5 kg head as one object with the arrow stuck in the head. With a closer look, I don't think that is accurate. The liner is like the liner in a hard hat. An arrow strike to the side might be metal against a thin piece of padded cloth. I don't know how to quantify that. It's probably around 1 to 2 cm space between the metal and head.

If the head and helmet is 8 kg together and the arrow get stuck(inelastic collision) it's:
m1 = 0.1 kg arrow
v1 = 48 m/s arrow (from the left)
m2 = 3 kg helmet + 5 kg head
v2 = 9 m/s (from the right)

(m1*v1 + m2*v2)/(m1 + m2)
(0.1*48 + 8*-9)/(0.1 + 8) = -8.2963 m/s

With that I can calculate the KE of the helmet (and arrow) moving to the left.
0.5*(m1+m2)*-8.2963^2
0.5*(0.1+8)*-8.2963^2 = 278.7558 Joule

The kinetic energy of the arrow and helmet together is:
(0.5*m1*v1^2)+(0.5*m2*v2^2)
(0.5*0.1*48^2)+(0.5*8*9^2) = 439.2 Joule439.2 - 278.7558 = 160.44 Joule going into the head and helmet.

160.44 - 80 = 80.44 Joule going into the head.

The KE of the arrow was 115.2 Joule, so the helmet and head moving at 9 m/s contributed:

160.44-115.2 = 45.24 Joule to the penetration. 28.2 %

Was that better?

 

[anorlunda]

[Moderator note: thread moved again to General Engineering]

 

[kuruman]

I think that is just to explain the 9m/s.

Yes, but if, as seems to be the case, the head is attached to the torso of a knight in armor firmly astride a galloping warhorse, should one not include the masses of all of the above when considering momentum conservation?

 

[256bits]

Yes, but if, as seems to be the case, the head is attached to the torso of a knight in armor firmly astride a galloping warhorse, should one not include the masses of all of the above when considering momentum conservation?

possibly.
The mentioned long bow was used to sling a hail of arrows into the air and coming down from above the enemy. 150 ft/sec could have been the total velocity of the arrow at something closer to between 30 and 60 degree elevation and not the horizontal Vx. Cross bow for short range and more accuracy, and long bow for "carpet bombing" the enemy. A knights head, or body, struck by an arrow coming from above would have less of a give, being pushed towards the ground or saddle.

This, of course complicates the calculation due to the angle of contact - ie not in a straight line.

Perhaps the 48 m/s is already the calculated given horizontal velocity against the 9 m/s knight velocity, afterall.
Which gives a reduced penetration power calculation of the arrow, as in this case Vtotal of arrow is something greater.

 

[haruspex]

Yes, but if, as seems to be the case, the head is attached to the torso of a knight in armor firmly astride a galloping warhorse, should one not include the masses of all of the above when considering momentum conservation?

No, that is hardly a rigid body. The neck would flex back. However, @256bits makes a good point about the angle.

 

[haruspex]

Was that better?

I asked you not to plug any numbers in until the very end.

 

[PeroK]

No, that is hardly a rigid body. The neck would flex back. However, @256bits makes a good point about the angle.

Perhaps, but if something hits a rider in the head, then he's likely to be knocked off the horse. You could exclude the horse, but an armour-clad neck might not flex very much. The significant flex might be in the hips.

 

[haruspex]

Perhaps, but if something hits a rider in the head, then he's likely to be knocked off the horse. You could exclude the horse, but an armour-clad neck might not flex very much. The significant flex might be in the hips.

Yes, but the flex would be progressive. In the first stage, whatever gap there is between helmet and head closes; next, the neck flexes a little, then progressively down the torso. Meanwhile, the arrow is making its way through the helmet. Hard to say at what stage the arrow makes it through.

 

[Warbow]

48 m/s is at 55 meters. Point blank the velocity of the arrow is 54.7 m/s. Mark Stretton shot a 102 gram arrow over a chronograph at 54.7 m/s with a 140 lb high alpine yew longbow.

Simon Stanley shot a 95.9 gram arrow at 53.3 m/s with a 150 lb Oregon yew bow. The arrow retained 43.3 m/s at ground level at 249.9 meters with 9 m/s tail wind.

That's my data points.

 

[Steelwolf]

There are several points to consider here, and since I used to build swords and armor for real, I have some points.

1. Most of the armor of the time was cuir bouli, or boiled leather which had hardness similar to hard plastics today. This would have been reinforced with iron strips or scales. This is from the type of armor you describe, lorica with lamina. While the metal can be punched thru, as well as the leather, the leather will have a pinching effect on the arrow, slowing it down.

2. The liner would have been multiple layer cloth or coiled rope, in either case giving a large mass of spongy fiber to go through which would have also caused the 'pinch effect'.

3. The English Longbow, while it certainly could be used for the longer distance arched shot, mostly during massed volley, but they were also particularly deadly and came close to crossbow-like power when used on the flat in 'close quarters'.

I know that in one battle the Welsh nailed the English Leader to his saddle by putting an arrow through his leg on one side, through the armor, leg and into the saddle, he turned to retreat and was nailed in the other leg in a similar way: through the armor, the leg and into the saddle, both arrows injured the horse beneath as well. There are war-points still poking through the inside of a Church door there, being built of heavy oak timbers.

So that bow was a deadly critter, but the range of types of armor and how the materials act is not straight forward, the materials used for the armor and the padding beneath it both make big differences, as well as if the arrow is a broadpoint or bodkin in style.

In essence there are too many variables. Armor then was not all of a kind, and metal was actually scarce, so it is not like being able to have a standard thickness or gauge of specific alloy material, or how it was work hardened or annealed in it's making. In Japan and China they found that heavy silk shirts could help minimize arrow wounds, so it was ordered that all soldiers wear them.

But again, I think there are too many variables as your question presently stands for there to be a meaningful answer unless you could account for the actual armor used and do actual trial testing. That would be a lot of fun, to my thinking.

This is also an example of as armor became stronger, the weapons became stronger to match. The English Longbow, and the soon to come gunpowder, put the heavily armored knights on horses out of business for the most part, and the battle you describe was a part of that story ladder.

 

[Steelwolf]

Re-reading your question, aside from the cuir bouli, the plate armor of the time was really not that thick and if it was Hard, it was brittle and prone to breaking, if it was tough then it was softer and prone to being malleable, allowing an arrow to penetrate by deforming the metal, in the hard metal it is possible to break a plate. Metallurgy tech of the era was not homogenous as it is today, and it was a constant battle between armor and weapon which has driven much of that technology, even today.

As it is, the materials involved would have been soft iron by our reckoning, for armor, and the tradeoff is always weight for protection VS mobility. Horseback one can carry more mass, but the head and arms still have to move. If the helmet is too heavy it can throw the balance of the fighter off, pulling him to his feet or off the horse. (the first helmet I made for SCA combat was too heavy, a 12 ga bascinet for a 6'6" gent, and the person whom it was custom ordered by, exactly to his specs, was not usable due to weight pulling him off balance, so that was a mutual learning experience that I still use as a fire pit {and keep the helm fires burning}) So there is a limit to the amount of material in metal used, and the padding is a huge factor in the pentration of an arrow. How much 'give' did the padding afford before the arrow pushed through the metal, did it push through or get captured by the liner?