Who can answer this technical conundrum and apparent paradox re ft/lb energy?

[andrewM]

Here is a question:

Say I have a pellet of 15g (grain). Its velocity is 550 ft per second in my air rifle. Using the calculator on the website below, this produces an energy level of 10.08 ft/lb.

https://www.airgundepot.com/airgun-calculators.html

Simple logic therefore suggests that each 55 ft/second velocity is equivalent to approx 1 ft/lb energy (550 ft sec divided by 10.08 ft/lb). Similarly, 5ft/lb energy would be 275 ft/second velocity (half of 550 ft/sec).

However, neither of these deductions is correct. Type in 55ft/second and the calculator produces a ft/lb energy of merely 0.1 ft/lbs. Type in 275 ft/sec and the calculator says the energy is only 2.52 ft/lbs, when it might have been expected to equate to 5 ft/lbs.

At the other end of the scale, if we raise the energy by merely 11 ft/second from 550 to 561 ft/sec, then the energy rises to nearly 10.5 ft/lbs from 10.08 ft/lbs.

The question is, why does this calculation not work equally; why does it work exponentially as the velocity increases, leading to a much faster rise in the ft/lb rate as the level of velocity rises?

It means that a rise of a few foot per second can significantly raise the ft/lbs rate. Thus, a difference of 11 ft per second velocity between the fastest and slowest pellet might seem well within the parameters of an efficient and consistent spring gun but this will also result in a rise of the ft/lb measure of nearly 0.5 ft/lb.

There is doubtless a simple answer; I wonder if one of the many experts here could kindly explain it.

Regards to all,
Andrew.

[bill57]

The relationship between kinetic energy and velocity is non-linear. Kinetic energy increases with the square of the velocity, KE=1/2 mv² (where m=mass of pellet, v=velocity of pellet)

Your units need to be Kg and Metres, and the answer will be in Joules. So roughly, KE=1/2 x 0.001 x 167², = 1/2 x 0.001 x 27889, = 13.9445 Joules = 10.28 ft/lb

[bullbarrel]

As the chap above said the velocity is squared in calculation (from memory)

ft/lbs = (velocity fps x velocity fps x pellet grs) / 450240.

So double the speed and 4 times the ft/lbs.

[Airsporterman]

I would see it as the energy driving the pellet is tailing off over it's flight, ie, it's starting at 10 lb/ft when it leaves the muzzle but by the time it reaches the end of it's flight, it's producing far less than that! Therefore it's not consistent for a number of factors, ie, drag/resistance, gravity, etc.
If I misunderstood, I apologise, my mind isn't what it wasn't!

ASM

[bill57]

I would see it as the energy driving the pellet is tailing off over it's flight, ie, it's starting at 10 lb/ft when it leaves the muzzle but by the time it reaches the end of it's flight, it's producing far less than that! Therefore it's not consistent for a number of factors, ie, drag/resistance, gravity, etc.
If I misunderstood, I apologise, my mind isn't what it wasn't!

ASM

I think my physics teacher would have said "Your observations are correct, but they're not the answer". Or as Mr Scott would say, "Ye cannae change the laws of Physics, Captain."
The relationship is defined by the equation KE=1/2mv² and is quite incontrovertible.

What you're talking about is how the pellet loses velocity, and thus energy, as it travels away from the muzzle. There are multiple factors involved in this, such as initial velocity, pellet mass, shape, frontal area, air density etc, and is an area that provokes continual discussion in all aspects of shooting.

[andrewM]

Thank you, all, for your learned responses. I have almost entirely forgotten what I learned in physics at school. Bill, can you explain, however, why kinetic energy and velocity is non-linear, in layman's terms? Thus, 10 ft/sec is nothing in itself but when added to a projectile already travelling at 550 ft/sec, it adds nearly 0.5 ft/lbs.

Rgds
A

[rancidtom]

Thank you, all, for your learned responses. I have almost entirely forgotten what I learned in physics at school. Bill, can you explain, however, why kinetic energy and velocity is non-linear, in layman's terms? Thus, 10 ft/sec is nothing in itself but when added to a projectile already travelling at 550 ft/sec, it adds nearly 0.5 ft/lbs.

Rgds
A

OK I'll try to simplify things and show the stages of working.
Kinetic Energy (in our case we refer to Muzzle Energy) = 1/2 M x V² where M is the mass of the object and V is the velocity of the object.
(For the present moment we will forget about units and conversion factors)
If we take a projectile of 1 unit mass travelling at 10 units of distance per second it will have a muzzle energy of
ME = 1/2 x ( 1 x 10²)
ME = 1/2 x ( 1 x 10 x 10)
ME = 1/2 x ( 1 x 100)
ME = 1/2 x 100
ME = 50

Now if we make the projectile travel at 12 units per second
ME = 1/2 x ( 1 x 12²)
ME = 1/2 x ( 1 x 144)
ME = 72
We can see there is a difference here of 22 units of energy when we go from 10 to 12 on our velocity reading. (72-50)

If we now look at a higher velocity, say 100 units per second:
ME = 1/2 x ( 1 x 100²)
ME = 1/2 x 10000
ME = 5000
and if we also look at 102 units per second:
ME = 1/2 x ( 1 x 102²)
ME = 1/2 x ( 1 x 104040)
ME = 5202
Now we have a difference of 5202 - 5000 which gives us a 202 unit difference in kinetic energy.
This is due to what we call in maths a "square law", it's not a "linear" relationship.

[bill57]

Can you explain, however, why kinetic energy and velocity is non-linear, in layman's terms?

Sorry Andrew, I can't. But neither can most folk on the internet, as the explanation requires some calculus, which is not really layman's terms (and my calculus has long departed). The simplest I can say is that velocity is a vector (ie a quantity that has both magnitude and direction), whereas energy is a scalar (a quantity that has only magnitude). Since a vector and a scalar can never be equal (this is implicit, since one has direction and the other does not), the relationship can never be linear.

Perhaps in layman's terms, analogy is better. This is the science behind 20mph rather than 30mph zones round schools, and those unsettling adverts where the child says, "If you hit me at 40mph, there's an 80% chance I'll die, but if you hit me at 30mph, there's an 80% chance I'll live". If a car strikes a pedestrian at the lower speed, the force of the impact is hugely reduced because of this non-linear relationship.

[Turnup]

Thank you, all, for your learned responses. I have almost entirely forgotten what I learned in physics at school. Bill, can you explain, however, why kinetic energy and velocity is non-linear, in layman's terms? Thus, 10 ft/sec is nothing in itself but when added to a projectile already travelling at 550 ft/sec, it adds nearly 0.5 ft/lbs.

Rgds
A

On a technicality KE is not measured in ft/lbs (pronounced as "feet per pounds" in the same way that ft/sec is pronounced "feet per second") - the unit is ft-lb or ft.lb, pronounced "foot pound". A certain Airgun magazine has got this wrong over a long period of time. It is easier to understand things if you use the correct terminology, and confusing or downright misleading if you don't

Back to the thread.....

Energy is an expression of the capacity to do work. Work is force times distance (note that KE is measured in foot-lbs which is a mathematical way of stating feet times pounds - force times distance). The same force acting over twice the distance is twice the work or twice the energy.

So a moving mass can exert a force against something trying to slow it down - that is to say the moving mass is giving up some of it's energy into whatever is trying to slow it down. If we allow it to slow to a halt then it has given up all of it's energy and the amount of force it took multiplied by the distance it has had to cover is the amount of work it has done, and this is the kinetic energy it has given up.

Taking your example of 10 ft/sec vs. 550 ft/sec let us imagine we apply a slowing force to reduce the projectile's velocity by 10 ft/sec. (i.e. 10 to zero or 550 to 540).

We can choose the force such that it takes one second to reduce the projectile's velocity by 10 ft/sec. We need not bother to calculate that force for the purposes of this explanation provided we can agree that such a force is indeed possible.

For the same projectile, the force needed to do this will be the same in both cases.

In both cases the energy change is the force used times the distance covered.

In the first case the distance covered in that one second is less than 10 feet (if it were not slowing down at all it could only cover at most 10 feet in one second).

In the second case the distance covered is at least 540 feet (since it would travel 550 feet in one second if not slowing down at all and 540 feet in one second if moving at it's slowest velocity)

So in the two cases the force applied is the same, the duration of that force is the same (one second) but the distances covered in that one second are very different even if we take the best possible distance (highest possible energy) in case 1 as 10 feet and the worst possible distance (lowest possible energy) in case 2 as 540 feet this is 54 times more energy.

Which is about as non mathematical as I can make it.

For the more mathematically minded, the actual distances are 5 feet and 545 feet so the energy in the second case is in fact 109 times the energy in the first case.