Adam
- 65
- 1
I'd like some comments on some things mentioned in another forum, if you would. Thanks.
Re: The physics of armour
I'm not a ballistics engineer, but isn't it just relative velocity that matters? The momentum of the target doesn't have anything to do with the ability of the projectile to penetrate as I remember from basic mechanics. Physics works within whatever inertial coordinate frame you choose to name. If you set your coordinate frame on the target, the target is effectively stationary. What MATTERS is the relative velocity of the projectile. The chance of penetration is the same however you define the coordinate system, therefore the determining factor is relative velocity, NOT momentum.
Remember that anywhere you choose on Earth includes the consideration of relative velocity as we are all moving with the Earth relative to the solar coordinate system (and it relative to the galaxy, and so on).
Take a look at published ballistics experiments... ALL of them base there results on the relative velocity of the projectile as seen from the target.
Therefore, if an arrow moves about 80 m/s and the target about 2 m/s (more realstic numbers), the contribution of the target to the relative velocity is about 2.5%. Pretty insignificant. A charging horse (say 8-10 m/s) would be more significant, but probably not enough to change the penetration characteristics terribly.
The other items you mention (elasticity, angle of incidence) all matter as well of course.
By the way... the values you calculated are not forces as far as I can tell. You calculated the momentum of the object (M=mv) instead of a Force (F=ma). In order for there to be forces invloved, they must have an acceleration. Since you stated the objects as having a steady state velocity, they are not accelerating, therefore the force on them before impact is zero (neglecting drag).
Of course, you meant KINETIC ENERGY as you stated later. the equation for this is KE=1/2mv^2 and the units of measure are Joules, not Newtons.
Precisely. Kinetic energy of the arrow as derived from its mass and the relative velocity between it and the target is what counts. "Momentum", "Power" and "Force" are not what one would use in this application. Friction is nearly irrelavent, and even so, is affected by "weather" in the most tenuous way (may as well add in "lunar tidal forces"). Additionally, metal plates used in armor are not "inelastic" - they do flex and deform on impact, and in late medieval times were often steels with a fairly high carbon content - close to what might be termed springsteel today.
For making such simple errors, I have and continue to question Mr. Weber's grasp of physics and wish he'd trouble himself to learn it before publicly expounding on the subject. (sa-lam!)
You can do basic physics on the subject relatively easily, but to do even the most basic ballpark analysis of what goes into the impact itself, I'm betting you'd need to run computer models using finite element analysis. To get reasonably correct results you need a computer simulation using something called "Hydrocode". That's something for the specialist engineers and PHDs.
In the end, it'd be easier to test resistance to arrows by taking a remotely triggered calibrated crossbow and a radar speedometer and stooting up assorted steel plates. Wear protection, think safety, and if you get hurt - it wasn't my idea!
Not exactly. Consider a similar example. You are running along at 10m/s. A 10' cube of steel is moving toward you, opposite direction, at 1m/s. Do you think the energy from that steel cube will be irrelevent, even though it contributes so little to the relative velocity of the impact? No. This is why mass matters.Therefore, if an arrow moves about 80 m/s and the target about 2 m/s (more realstic numbers), the contribution of the target to the relative velocity is about 2.5%. Pretty insignificant. A charging horse (say 8-10 m/s) would be more significant, but probably not enough to change the penetration characteristics terribly.
The other items you mention (elasticity, angle of incidence) all matter as well of course.
Yeah, I'd be really interested if someone could find such figures for me as would apply to this situation.
By the way... the values you calculated are not forces as far as I can tell. You calculated the momentum of the object (M=mv) instead of a Force (F=ma).
Force is Force. A nice explanation from the web:
For a body of mass m moving at non-relativistic velocity v, the momentum is mv. The force, F, is then defined as d(mv)/dt. If the mass is constant, F = m dv/dt = ma where a is the acceleration.
The "a", or acceleration, is actually what we call negative acceleration in the case of an impact such as this. Some people have the habit of saying "minus fifteen Newtons" or such, but this is incorrect.
In order for there to be forces invloved, they must have an acceleration. Since you stated the objects as having a steady state velocity, they are not accelerating, therefore the force on them before impact is zero (neglecting drag).
The (negative) acceleration occurs at impact, which is why we can say there is Force.
Of course, you meant KINETIC ENERGY as you stated later. the equation for this is KE=1/2mv^2 and the units of measure are Joules, not Newtons.
No, you're got that wrong, sorry. Joules is a measure of energy, yes, but tells us absolutely nothing about what is happening. A can of Coke sitting on the bench, doing nothing, has a certain number of Joules. A boulder at the top of a hill, doing nothing, has Joules. It has those Joules regardless of wether it is rolling down the hill, sitting there whistling Dixie, or playing cards. It tells us nothing about what is happening. This is why it is incorrect to state the results of impacts in Joules. It's like stating "I weigh 90 kg" or such. Everyone does it, but it's wrong. Weight should be measured in Newtons. To find out what is actually happening, you need to state Watts (telling us the rate at which energy is being used up), or Newtons (telling us the Force of the impact).
In other words... the KE formula tells us how much energy is involved in getting the arrow moving. KE=1/2mv^2 using example figures of 20 grammes and 50m/s gives us, for example KE= 10g*2,500 = 25,000 Joules. However, the arrow has that energy regardless of whether it hits anything or not. In fact it loses that energy very swiftly as it travels. The amount of energy tells us absolutely nothing about the impact. Thus... Force, in Newtons.
Consider a similar example. You are fall off your chair, hitting the ground at say 2 m/s. Due to the time of day, it so happens that the Earth is moving toward you, opposite direction, at 144,000 m/s (speed taken from Monty Python's Galaxy song). Do you think the energy from that planet will be relevant, even though the relative velocity of the impact is so small?
Drag (friction against the air) reduces arrow speed from about 80m/s to around 50m/s after roughly a hundred metres. Quite significant.Friction is nearly irrelavent, and even so, is affected by "weather" in the most tenuous way (may as well add in "lunar tidal forces").
Additionally, metal plates used in armor are not "inelastic" - they do flex and deform on impact, and in late medieval times were often steels with a fairly high carbon content - close to what might be termed springsteel today.
In terms of physics, it is an inelastic collision.