How is energy dissipated in a nail gun?
How is energy dissipated in a nail gun?
(OP)
Hi all,
I'm having trouble working out how much energy will have to be dissipated in a test rig for a nail gun.
From previous testing, I know that a test slug fired from the gun has a mean energy of 100J.
I now want to design a test rig which can fire the gun repeatedly for 100,000 shots.
I have decided that firing a test slug vertically upwards in to the bottom of a heavy piston in a steel tube will be the most suitable design.
Each time the slug is fired from the gun, it collides with a rubber-bottomed piston in a steel tube, thus transferring its energy to the piston by pushing it up the tube. The piston can then bounce up and down in the tube until its energy is entirely dissipated as heat and sound. The process is then repeated 100,000 times.
The problem I have with designing the rig, is working out how high the piston will travel within the tube. I know I'm making a flawed assumption somewhere so please let me know where my flawed reasoning is!
Here is my working and reasoning:
Assumptions:
- neglect energy dissipation due to air drag
- since the rubber bottom of the piston has a very high spring constant, assume that once the piston and slug collide, the slug and the piston both travel with the same velocity
m_s = mass of the test slug = 0.09 kg
m_p = mass of the piston = 10 kg
E_i = energy of the test slug exiting the gun = 100 J
E_f = combined kinetic energy of the test slug and piston immediately after the collision
v_s = velocity of the test slug before the collision
v_f = velocity of both the slug and the piston immediately after the collision
Working out the velocity of the test slug:
v_s = sqrt(2*E_i / m_s) (from the kinetic energy equation)
= sqrt(2*100 / 0.09)
= 47.1 m/s
Using the conservation of momentum principle, determine the velocity of the slug and piston immediately after the collision:
m_s * v_s = (m_s + m_p) * v_f
v_f = m_s/(m_s + m_p) * v_s
= 0.09/(0.09 + 10) * 47.1
= 0.42 m/s
Combined kinetic energy of the slug and piston immediately after the collision:
E_f = 0.5 * (m_s + m_p) * v_f^2
= 0.5 * 10.09 * 0.42^2
= 0.9 J
Thus the maximum height that the piston will reach:
h = E_f / ((m_s + m_p) * g)
= 0.9 / (10.09 * 9.81)
= 9 mm
I then double checked the maximum height of the piston using the conservation of energy equation and the initial energy of the test slug. I assumed that the kinetic energy of the test slug would be entirely converted to gravitational potential energy of the piston once the piston reached its highest point. Thus energy losses due to sound and drag etc were neglected.
h = E_i / ((m_s + m_p) * g)
= 100 / (10.09 * 9.81)
= 1 metre
Thus the maximum height differs significantly depending on the calculation method I use.
Is my conservation of momentum approach correct?
Is my conservation of energy approach correct?
My intuition tells me that the conservation of momentum approach is correct, however, I'm suspicious since I do not believe that the kinetic energy of the system would be reduced from 100J to 0.9J from the collision (i.e. huge energy losses to sound and heat).
Please help!
Thanks,
Paul
I'm having trouble working out how much energy will have to be dissipated in a test rig for a nail gun.
From previous testing, I know that a test slug fired from the gun has a mean energy of 100J.
I now want to design a test rig which can fire the gun repeatedly for 100,000 shots.
I have decided that firing a test slug vertically upwards in to the bottom of a heavy piston in a steel tube will be the most suitable design.
Each time the slug is fired from the gun, it collides with a rubber-bottomed piston in a steel tube, thus transferring its energy to the piston by pushing it up the tube. The piston can then bounce up and down in the tube until its energy is entirely dissipated as heat and sound. The process is then repeated 100,000 times.
The problem I have with designing the rig, is working out how high the piston will travel within the tube. I know I'm making a flawed assumption somewhere so please let me know where my flawed reasoning is!
Here is my working and reasoning:
Assumptions:
- neglect energy dissipation due to air drag
- since the rubber bottom of the piston has a very high spring constant, assume that once the piston and slug collide, the slug and the piston both travel with the same velocity
m_s = mass of the test slug = 0.09 kg
m_p = mass of the piston = 10 kg
E_i = energy of the test slug exiting the gun = 100 J
E_f = combined kinetic energy of the test slug and piston immediately after the collision
v_s = velocity of the test slug before the collision
v_f = velocity of both the slug and the piston immediately after the collision
Working out the velocity of the test slug:
v_s = sqrt(2*E_i / m_s) (from the kinetic energy equation)
= sqrt(2*100 / 0.09)
= 47.1 m/s
Using the conservation of momentum principle, determine the velocity of the slug and piston immediately after the collision:
m_s * v_s = (m_s + m_p) * v_f
v_f = m_s/(m_s + m_p) * v_s
= 0.09/(0.09 + 10) * 47.1
= 0.42 m/s
Combined kinetic energy of the slug and piston immediately after the collision:
E_f = 0.5 * (m_s + m_p) * v_f^2
= 0.5 * 10.09 * 0.42^2
= 0.9 J
Thus the maximum height that the piston will reach:
h = E_f / ((m_s + m_p) * g)
= 0.9 / (10.09 * 9.81)
= 9 mm
I then double checked the maximum height of the piston using the conservation of energy equation and the initial energy of the test slug. I assumed that the kinetic energy of the test slug would be entirely converted to gravitational potential energy of the piston once the piston reached its highest point. Thus energy losses due to sound and drag etc were neglected.
h = E_i / ((m_s + m_p) * g)
= 100 / (10.09 * 9.81)
= 1 metre
Thus the maximum height differs significantly depending on the calculation method I use.
Is my conservation of momentum approach correct?
Is my conservation of energy approach correct?
My intuition tells me that the conservation of momentum approach is correct, however, I'm suspicious since I do not believe that the kinetic energy of the system would be reduced from 100J to 0.9J from the collision (i.e. huge energy losses to sound and heat).
Please help!
Thanks,
Paul





RE: How is energy dissipated in a nail gun?
TTFN
I can do absolutely anything. I'm an expert!
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers
RE: How is energy dissipated in a nail gun?
If the test slug was free to rise without hitting a piston, how high would it rise? That would fit the conservation of energy approach.
Ted
RE: How is energy dissipated in a nail gun?
What is the point of the test and why do you want to capture the energy?
RE: How is energy dissipated in a nail gun?
If the objective is to cycle the nail gun 100,000 times presumably without some failure, then I would choose to shoot the nails downward into a barrel that has a few inches, cm, of sand in the bottom to prevent nail rebound and to collect the nails.
Ted
RE: How is energy dissipated in a nail gun?
RE: How is energy dissipated in a nail gun?
100J is not much heat.
hydtools solution looks good.
je suis charlie
RE: How is energy dissipated in a nail gun?
I don't think I'd expect the bumpers inside the nail gun would survive 100,000 "dry" firings very well.
RE: How is energy dissipated in a nail gun?
It might be that a relatively uniform piece of wood target, perhaps fir, and an engineering intern to fire the gun for 100,000 cycles would work.
Ted
RE: How is energy dissipated in a nail gun?
Apologies for the late reply to some of your questions.
The objective of the test is purely to determine if and how the gun may fail before having fired 100,000 nails. I want it to be as realistic as possible in that I would like the test slug to be slowed to a standstill over 90 mm (which is the length of a standard nail).
I've considered firing nails in to wood but this would cost too much (the cost of 100,000 nails) and may require too much human input to replace the wood and complexity in that the gun or wood would have to move between each shot. The sand box is a good idea as I could recover the nails; however, it might be difficult to make the conditions realistic. Also, that's a really good idea having an engineering intern fire the gun - a classic intern project! Although the cost of nails would be too high unfortunately.
I have therefore opted to use a 90mm long pin which screws in to a flexible rubber isolator which then screws in to a 2.5kg 'piston'. Thus when the gun fires, its striking pin hits the 90mm long pin (which fits inside the gun's barrel where the nail would sit) which causes the 90mm pin/isolator/'piston' to travel 90mm vertically up inside a seamless steel pipe. The pin/isolator/'piston' assembly then falls back in to the gun's barrel (since it is guided by a lead-in) and bounces until its energy has been dissipated as heat.
I have allowed for the 'piston's weight to be adjustable to account for the assumptions that I've used in the conservation of momentum equation. I will also check the temperature of the assembly to ensure that it isn't getting too hot.
What do you think of this solution?
RE: How is energy dissipated in a nail gun?
Bouncing implies time, so it's not clear that this is much of a savings
Re-using the same nail seems dubious; there's bound to be some wear and tear on the nail, which might cause jamming.
TTFN
I can do absolutely anything. I'm an expert!
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers
RE: How is energy dissipated in a nail gun?
RE: How is energy dissipated in a nail gun?
Buy the nails and wood and get it done.
Ted
RE: How is energy dissipated in a nail gun?
Ted
RE: How is energy dissipated in a nail gun?
However, the clients would only like to spend a few thousand on a rig which can complete several tests and may also be used for testing a concrete gun prototype.
Thus I think that the cost of nails makes this approach unfeasible (especially for the concrete gun). I like the sandbox approach where I could reuse the nails; however, I can't see how it would provide a more realistic scenario than that which I've come up with (i.e. the sand wouldn't provide adequate resistance as far as I can see).
For my approach, I agree that the recoil and other characteristics will be different; however, by slowing the pin (which is simulating the nail) over the same distance which a nail is driven in to wood (90mm), it will be as close to reality as I can get it given the constraints of the project.
With this extra information (that I have to do several tests and also on a concrete gun for which the nails cost $0.56 each), do you have any suggestions for a better testing method than what I've chosen?
RE: How is energy dissipated in a nail gun?
RE: How is energy dissipated in a nail gun?
je suis charlie
RE: How is energy dissipated in a nail gun?
Barely, at best. The spring constant of the rubber is very different than that of wood. Wood doesn't really spring back. Perhaps something thixotropic?
TTFN
I can do absolutely anything. I'm an expert!
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers
RE: How is energy dissipated in a nail gun?
Ted
RE: How is energy dissipated in a nail gun?
A.
RE: How is energy dissipated in a nail gun?
I’ll bet the nail gun manuf’er. and the nail manuf’er. don’t think nails are too expensive when the poor carpenter trying to make a living has to buy their proprietary nails which work only in their gun. I don’t think you’ll be testing the gun accurately and truthfully unless you fire regular collated nails into wood. If you want to design something mechanical design a system which indexes the 2x wooden members under the gun btwn. each firing. Or index the gun on a gun holding system, up and down the wooden members, at regular intervals, btwn. firings. Driving real nails into wooden members (varying btwn. oak, pine, SYP and DF) causes the real world forces and reactions on the gun, to test it: and will also reveal nail jams, bending, breakage, etc. which are problems which the rest of us hope are resolved before we pay a premium for your nail gun.
RE: How is energy dissipated in a nail gun?
RE: How is energy dissipated in a nail gun?
I doubt the nail gun will successfully complete the first test run without failure. So the consumables will have to be in such quantity to eventually complete the last 100,000 cycles without failure.
Ted
RE: How is energy dissipated in a nail gun?
Some other musings:
I'd bet most nail guns spend their lives breathing very wet compressed air. Bone dry shop air won't reflect the real world. This assumes a pneumatic gun, of course.
I'd bet most nail gun mechanisms are infiltrated by some significant amount of abrasive grit. They live in the back of a work truck, on a construction site. A clean shop environment won't reflect this either.
I'm not sure about reusing nails; all the nail gun "ammo" I've ever seen is joined by wire or paper or something. I don't think you can feed a nail gun loose nails.
If the client only wants to spend "a few thousand", the case is even stronger for heading to Home Depot for some wood and nails, not building a machine. A machine would surely look fancy and scientific, but the real world wood/nails marathon could be just as scientifically rigorous. Truly superb data could be gathered, say photos and measurements taken during a full tear down every 10K nails, or whatever. If I were the nail gun product engineer, I'd be salivating over data like this.
However, if the goal is to construct a test stand to COMPARE different nail gun designs...lubricants...maintenance schedules...whatever, THEN I could see the value of building such a machine. Even then, "a few thousand" is wishful thinking.