Page 1 of 3 123 LastLast
Results 1 to 15 of 31

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

  1. #1
    Join Date
    Jan 2017
    Location
    Pulborough
    Posts
    997

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

    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.

  2. #2
    Join Date
    Jan 2018
    Location
    Glasgow
    Posts
    925
    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
    Last edited by bill57; 19-02-2019 at 11:48 PM.

  3. #3
    Join Date
    Nov 2005
    Location
    Gone West Young Man
    Posts
    20,269
    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.

  4. #4
    Airsporterman's Avatar
    Airsporterman is offline Makes Scrooge look Happy and Generous!
    Join Date
    Nov 2010
    Location
    Moving target, nr Blyth, God's Northumberland
    Posts
    18,970
    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 am a Man of La Northumberlandia, a true Knight and spend my days on my Quest (my duty nay privilege!) and fighting dragons and unbeatable foe, to right the unrightable wrongs, to bear with unbearable sorrow and dreaming my impossible dreams.

  5. #5
    Join Date
    Jan 2018
    Location
    Glasgow
    Posts
    925
    Quote Originally Posted by Airsporterman View Post
    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.
    Last edited by bill57; 20-02-2019 at 12:20 AM.

  6. #6
    Join Date
    Jan 2017
    Location
    Pulborough
    Posts
    997
    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

  7. #7
    Join Date
    Jul 2009
    Location
    Manchester
    Posts
    1,098
    Quote Originally Posted by andrewM View Post
    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.
    BSA Super10 addict, other BSA's inc GoldstarSE, Original (Diana) Mod75's, Diana Mod5, HW80's, SAM 11K... All sorted!

  8. #8
    Join Date
    Sep 2010
    Location
    Bruton
    Posts
    6,616
    https://en.m.wikipedia.org/wiki/Kinetic_energy

    Or try this.

    I have no idea why this is the case (non-linear relationships between increases in mass versus increases in velocity) but physics says that similar increases in velocity at the same weight increase energy far more than similar increases in mass while velocity stays constant.

    Momentum, on the other hand, is linear.

    https://en.m.wikipedia.org/wiki/Momentum

    I hate physics. It’s annoying. Unfortunately, it appears to be real and true. Bloody science.

  9. #9
    Join Date
    Jan 2018
    Location
    Glasgow
    Posts
    925
    Quote Originally Posted by andrewM View Post
    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.

  10. #10
    Join Date
    Jan 2018
    Location
    Glasgow
    Posts
    925
    Quote Originally Posted by Geezer View Post
    I hate physics. It’s annoying. Unfortunately, it appears to be real and true. Bloody science.
    When I was at school, the standard text was "Physics is Fun" by Jim Jardine. Apparently it was not uncommon to open one, and pencilled inside would be "Jardine's a bloody liar".

  11. #11
    Join Date
    Sep 2010
    Location
    Bruton
    Posts
    6,616
    Quote Originally Posted by bill57 View Post
    When I was at school, the standard text was "Physics is Fun" by Jim Jardine. Apparently it was not uncommon to open one, and pencilled inside would be "Jardine's a bloody liar".
    The only time I can recall physics being fun was building a ballistic pendulum to test the muzzle energy of an ASI/Gamo Expo/Sniper (the school’s not mine) and then me explaining why it’s results were wrong. And, though not physics, being told that I could definitely not build a Sten gun as my metalwork project.

  12. #12
    Join Date
    Jun 2012
    Location
    North Wales
    Posts
    3,275
    In layman's terms. Try riding a bicycle on the flat in still air. Unless you're an accomplished cyclist getting above 15 mph, sitting upright, becomes very hard work. Non linear drag

  13. #13
    Join Date
    Jan 2017
    Location
    Pulborough
    Posts
    997
    How very kind of you scientific fellows to try to re-educate me and to take the time to explain the physics, here (this forum has some considerable expertise).

    I understand the formula and the calculation. Trying to rationalise it in my mind is rather less easy but, happily, I think I am not the only one, here! I am also better informed that when I began.

    Rgds to all.
    A

  14. #14
    ccdjg is offline Airgun Alchemist, Collector and Scribe
    Join Date
    Jun 2007
    Location
    Leeds
    Posts
    2,173
    At the risk of boring the pants off everyone, I can add a bit more to the explanation. It is a misconception to think of the kinetic energy of a moving object as coming directly from its velocity, as it actually comes directly from the acceleration needed bring the object up to that velocity in the first place.

    Imagine the pellet sitting in the barrel at rest, with zero velocity and zero energy. When the force from the compressed hits it, it is then rapidly accelerated along the barrel under the effect of the constantly applied force, the velocity increasing progressively until it leaves the muzzle. The law of physics states that the energy transferred to an object in such a process is given by the mass of the object, m, multiplied by the acceleration, a, and the distanced moved d.
    Easy to remember as: E = m.a.d.

    Acceleration is not the same as velocity, but is the rate at which the velocity is increasing. Acceleration times distance, a.d., is related to velocity squared, not velocity, which is why the kinetic energy of the pellet is not simply proportional to the velocity of the pellet.

  15. #15
    Turnup's Avatar
    Turnup is offline Dialling code‎: ‎01344
    Join Date
    Nov 2012
    Location
    Crowthorne
    Posts
    5,512

    Why is energy non linear with velocity - an attempt at an explanation without maths

    Quote Originally Posted by andrewM View Post
    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.
    True freedom includes the freedom to make mistakes or do foolish things and bear the consequences.
    TANSTAAFL

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •