the hammer's blow
Mark Finn
bounce ? put hot steel in the middle and work is done. all you
blacksmiths out there, did you know that when you strike your anvil
using a 40 ounce hammer, for a split second you have obtained the force
of 13 to 18 tons !!! believe me when i tell you there is real power in
blacksmithing.
have fun, mark
Martin H. Eastburn
large flat head spreads the force sideways and drops the pressure
on a unit area.
Ball peens are the blacksmith utility hammer - face and ball ability.
Then a wide flange and a pick on the other.
A 5 pound hammer with a flat point or knife edge is useful for cuts.
When supporting a company that had hundreds of engineers running
here and there getting product out and trying for 6 sigma - I worked
through a contract and po that defined 'no undetectable faults to be allowed
during high speed data transfer'. It had taken them a year and a half
to unpack the box with cables and boards. In weeks we were running 100%
on 40 machines. Management signed it off. My management was flabbergasted.
The salesman who allowed the crazy thought was long gone.
Mark was referring to a domed face. The center spot is tiny with a very
high pounds/in^2 since the area is reduced, the force increases.
Martin
Curt Welch
> Pounds per square inch. e.g. a pick will punch through while a
> large flat head spreads the force sideways and drops the pressure
> on a unit area.
>
> Ball peens are the blacksmith utility hammer - face and ball ability.
>
> Then a wide flange and a pick on the other.
>
> A 5 pound hammer with a flat point or knife edge is useful for cuts.
>
> When supporting a company that had hundreds of engineers running
> here and there getting product out and trying for 6 sigma - I worked
> through a contract and po that defined 'no undetectable faults to be
> allowed during high speed data transfer'. It had taken them a year and
> a half to unpack the box with cables and boards. In weeks we were
> running 100% on 40 machines. Management signed it off. My management
> was flabbergasted. The salesman who allowed the crazy thought was long
> gone.
Not sure what that paragraph has to do with hitting an anvil with a hammer.
:)
> Mark was referring to a domed face. The center spot is tiny with a very
> high pounds/in^2 since the area is reduced, the force increases.
Well, you are confusing force with pressure there. Force is like weight.
It has nothing to do with the area. Pressure however is force per unit
area. Mark only seemed to be talking about force.
The total force generated is a function of the elasticity or hardness of
the hammer and the anvil. The harder the material, the higher the force
generated when the hammer bounces. The harder the steel, the less it
compresses, and the faster it bounces. The faster it bounces, the more
force is generated to make the hammer change direction in the shorter
period of time.
A flat hammer with large contact area will change direction faster making
it generate more force than a domed hammer of the same weight and speed
even though the pressure (force per area) will be larger with the smaller
contact area. So the force actual goes down when you reduce the contact
area (something which is not at all intuitive becuase we do normally think
of pressure and not of force).
I have no clue what forces are really generated because I've seen no real
data on the subject, but it's easy enough to believe that a hammer hitting
a anvil will create many tons of force for the very short period of time
the two are in contact before the hammer bounces off the anvil.
I feel the need to do some math to see how the numbers work out...
I'll assume the hammer and anvil work like a linear spring when it bounces.
A 40 oz hammer is about 1 kg.
If it's moving at 5 m/s at impact (I guess that's a reasonable number) the
kinetic energy in the hammer is 1/2 m v^2 or 1/2 * 1 * 25 or about 12.5
Joules.
The kinetic energy in the hammer is converted to potential energy in the
compressed spring which causes the hammer to stop moving, then the process
is reversed, and the potential energy is returned to the hammer as it
bounces. Some energy is lost as heat in the hammer and anvil, but lets
just assume no energy is lost.
U = 1/2 kx^2 is the potential energy stored in the spring - which must
equal the potential kinetic energy in the hammer before impact. The
maximum force happens when all the kinetic energy is transferred to
potential energy and the hammer stops moving. K is the compression
constant for the spring, and x is the distance the spring moves as it's
compressed.
F = kx is the force produced by the spring as it's compressed a distance of
x.
Solving the first equation for kx, we get:
kx = 2U/x
so
F = 2U/x
Where U for our example is 12.5 Joules, and x is the distance the anvil
compresses.
We have to guess at x. The smaller x however, the higher the force. Lets
guess .025 mm (which is about 1/1000 of an inch), or .000025 m. That means
the force becomes:
F = 25/.000025 or 1,000,000 Newtons.
1 Newton is about .225 lb or .0001125 ton.
So 1,000,000 Newtons is 112.5 tons.
So these numbers (assuming I got the math right) produce an answer that is
10 times larger than what Mark was reporting. But it was just a wild guess
on my part to come up with the .025 mm amount of compression. That could
easily be off by a factor of 10.
Also note that these high force numbers only happen when you hit hard steel
to hard steel which results in very little compression and a very fast
bounce. The very high force numbers are generated in order to make the
hammer switch direction so quickly. Put a piece of soft hot metal in there
(or a soft finger) and the max force goes way down. If in this example, we
move the metal by 5 mm when we hit it, the max force becomes 25/.005 x
.0001125 or 0.56 tons or around 1000 lbs.
Hit a large lump of clay that is compressed by 100 mm at impact and the
force goes all the way down to 56 lbs. That's becuase the force has a much
larger distance and time to act on the hammer and make it slow down.
> Martin
>
> Mark Finn wrote:
> > what happens when an unstoppable force hits an unmoveable object ?
> > bounce ? put hot steel in the middle and work is done. all you
> > blacksmiths out there, did you know that when you strike your anvil
> > using a 40 ounce hammer, for a split second you have obtained the force
> > of 13 to 18 tons !!! believe me when i tell you there is real power in
> > blacksmithing.
> > have fun, mark
> >
--
Curt Welch http://CurtWelch.Com/
cu...@kcwc.com http://NewsReader.Com/
Mike Spencer
cu...@kcwc.com (Curt Welch) wrote:
> If it's moving at 5 m/s at impact (I guess that's a reasonable
> number)...
I think it's more like 10 m/s. See:
http://home.tallships.ca/mspencer/vec.html
based on some high-speed photography I did years ago.
See also: http://home.tallships.ca/mspencer/f-plot3.html
http://home.tallships.ca/mspencer/hammer.html
I've been a bit boggled trying to figure out how to calcualte the
forces and especially conservation of momentum involved in hammering
hot iron. Yes, I can do billiard balls, balistic pendulum and the like,
just not what happens on the anvil. (Maybe I just have to include the
mass of the earth (6e34 kg) in my figgering.)
Maybe that's the reason why, in engineering books, forging machinery is
rated by various practical tests such as striking a single blow on a
lead cylinder of specified dimensions and then measuring the
deformation.
Here's a page that might interest those for whom Curt's post was
just a tasty start:
http://www.animations.physics.unsw.edu.au/jw/momentum.html
--
Mike Spencer Nova Scotia, Canada
Curt Welch
> cu...@kcwc.com (Curt Welch) wrote:
>
> > If it's moving at 5 m/s at impact (I guess that's a reasonable
> > number)...
>
> I think it's more like 10 m/s. See:
>
> http://home.tallships.ca/mspencer/vec.html
Cool. Do you have any sense how hard that was being hit when it reached
1000 cm/sec? that is, would you say it was an easy, medium, or hard hit?
Was it you swinging the hammer?
> based on some high-speed photography I did years ago.
>
> See also: http://home.tallships.ca/mspencer/f-plot3.html
> http://home.tallships.ca/mspencer/hammer.html
>
> I've been a bit boggled trying to figure out how to calcualte the
> forces and especially conservation of momentum involved in hammering
> hot iron. Yes, I can do billiard balls, balistic pendulum and the like,
> just not what happens on the anvil. (Maybe I just have to include the
> mass of the earth (6e34 kg) in my figgering.)
Not sure why you would want to look at momentum. Sure, you have to have
take the earth into account if you look at momentum equations but then it
so heavily dominates the equations as to be pointless. When something hits
the earth, you just assume the earth has the power to suck up all the
momentum for any part of the collision which isn't elastic. I don't think
you can't calculate how an object will deform based on momentum equations.
It's a property of the material created by attraction and momentum effects
at the molecular level. If it could be calculated with physics equations,
it would have to be done at the molecular level (or some estimate near
that) I think.
> Maybe that's the reason why, in engineering books, forging machinery is
> rated by various practical tests such as striking a single blow on a
> lead cylinder of specified dimensions and then measuring the
> deformation.
Yeah, I think that would just have to be how it's done. You must measure
it empirically. Once you had a good model of the material, then I think it
could be calculated based on momentum and shape of the hammer head as well
as the shape (and of course the heat) of the steel being hit.
One thing I think would be interesting to understand about a hammer blow is
how the hammer is whipped in a typical hammer motion so as to transfer
energy that has probably been put into the arm and hammer handle, into the
hammer head right before impact. I would guess that the best type of
hammer swing will minimize shock to the body parts and instead transfer all
the energy in the body into the hammer head so there is no sudden shock to
the body (elbow, wrist, shoulder) for each impact. That would also I would
guess get maximum energy into the hammer head so all the energy you put
into the swing goes into the work, instead into your joints and bones.
In your plots, the hammer head accelerates, but did you calculate the
momentum of the handle or the person's arm, separate from the head? For an
efficient swing, I would expect to see the energy in the arm, and the
handle go up as you bring the hammer down, but right before impact, drop
quickly to near zero as the energy dumps into the hammer head.
Mark Finn
you guys. my background is more toward the arts. my experience forging
at the anvil and using a 50 ton forging press lead me to make the
comparison of moving hot steel. i've seen cold steel behave like clay or
plastic using the forging press. put a coin between two pieces of mild
steel at room temp. then apply say, 40 tons. although the coin is softer
it will leave it's impression in both pieces !
have fun, mark
Mike Spencer
cu...@kcwc.com (Curt Welch) wrote:
>Mike Spencer <m...@bogus.nodomain.nowhere> wrote:
>> cu...@kcwc.com (Curt Welch) wrote:
>>
>> > If it's moving at 5 m/s at impact (I guess that's a reasonable
>> > number)...
>>
>> I think it's more like 10 m/s. See:
>>
>> http://home.tallships.ca/mspencer/vec.html
>
> Cool. Do you have any sense how hard that was being hit when it reached
> 1000 cm/sec?
Good, solid forging blows, not wimping, not straining.
> Was it you swinging the hammer?
Yes. The sepia collage is me.
> Not sure why you would want to look at momentum.
Well, maybe it's a red herring. Muddling about with the math of
hammering is part of a (very gradual) project to write a graphic model
of a Little Giant-type power hammer sufficiently correct that the
typical bad behavior of a maladjusted hammer can be reproduced. With
that level of modeling, one should then be able to tweak the
parameters of the model to find correct adjustments for one's hammer.
> When something hits the earth, you just assume the earth has the
> power to suck up all the momentum for any part of the collision
> which isn't elastic.
So if you assume conservation of momentum, then you have to vanish the
momentum (of, say, a billiard ball falling into a tank of JellO) into
something like 10e-24 meters/s of the earth. Ho hum.
On a visit to MIT, I cornered one of the part-time faulty for the team
I was working with and asked him to answer some questions about KE and
mv. But he was so keen on reproducing his classroom presentation that
he couldn't listen to my questions.
Another prof I visited there was a robotics wizard, had a robot that
could play paddle ball and another that could catch a thrown tennis
ball. I wanted him to implement a hammering robot. Not one that
would have the power to forge actual hot iron, just one that could go
through the correct motions with, say, a piece of plasticene or
similar. He said it was "too hard". Huh.
> One thing I think would be interesting to understand about a hammer blow is
> how the hammer is whipped in a typical hammer motion so as to transfer
> energy that has probably been put into the arm and hammer handle, into the
> hammer head right before impact. I would guess that the best type of
> hammer swing will minimize shock to the body parts and instead transfer all
> the energy in the body into the hammer head so there is no sudden shock to
> the body (elbow, wrist, shoulder) for each impact. That would also I would
> guess get maximum energy into the hammer head so all the energy you put
> into the swing goes into the work, instead into your joints and bones.
>
> In your plots, the hammer head accelerates, but did you calculate the
> momentum of the handle or the person's arm, separate from the head?
I think the most notable features of the flight-path plot are that the
hammer head moves in almost a straight line during all but a bit of
the first 2/3 of its flight and that it's essential rotating around
my hand in the last 1/5 or so. In the last few lines of the plot, the
hand (blue dot) doesn't move much and is mostly just providing the
centrepital force to hold the head on its path.
I don't recall if the frames in the collage were at equal time
intervals or not and it would take some digging to unearth the
original images to find out. You could assume so, take pixel
measurements from the "crash test dummy" marks on my shirt and so the
calculations yourself. :-)
> For an efficient swing, I would expect to see the energy in the arm,
> and the handle go up as you bring the hammer down, but right before
> impact, drop quickly to near zero as the energy dumps into the
> hammer head.
You can make some guesstimates about that from looking at the blue-dot
intervals in the flight plot.
- Mike
bigegg
> cu...@kcwc.com (Curt Welch) wrote:
>
>> Mike Spencer <m...@bogus.nodomain.nowhere> wrote:
>>> cu...@kcwc.com (Curt Welch) wrote:
>>>
>>>> If it's moving at 5 m/s at impact (I guess that's a reasonable
>>>> number)...
>>> I think it's more like 10 m/s. See:
>>>
>>> http://home.tallships.ca/mspencer/vec.html
>> Cool. Do you have any sense how hard that was being hit when it reached
>> 1000 cm/sec?
>
> Good, solid forging blows, not wimping, not straining.
I'd calculate it like this:
F=M x A
the hammer decelerates to zero at some point (just before it rebounds),
the length of time required depending upon the item being hit.
That brings me to the conclusion that the force of a hammer blow depends
upon the item being hit - an anvil which could be considered virtually
"immoveable" will be hit with (virtually) infinite force, because the
deceleration will be near instantaneous.
A better question might be what energy is imparted by a hammer blow -
this one is easy -
E = 0.5mvv
so for a 5kg (11lb) hammer at 10m/s = 500 J
Assume that this deforms a red hot piece of steel weighing 100g by
1cm (0.01m), and the hammer rebounds at 1/2 it's incoming speed
then outgoing kinetic energy = 250J leaving deformation energy as 250J
1J =1 kgm/s^2
250J = 0.1kg0.01m / s^2
s^2 = 0.1 x 0.01 / 250
s^2 = 0.001 / 250
s = 0.002 seconds
a 5kg hammer, hitting at 10m/s second and rebounding at 5 m/s 0.002
seconds later has an acceleration of 7500m/s^2 (15m/s / 0.002s)
and hits with a force of 5 x 7500m/s = 37500 N (or 3.75 Tonnes if you
prefer)
BUT -
there's a LOT of assumptions there -
the real answer is:
it depends.
--
bigegg
Curt Welch
Looks like you dropped one measure of distance.
that would be 1 kg m^2 / s^2
F=ma and energy = Fd so
Energy = mass * (distance / second^2) * distance
> 250J = 0.1kg0.01m / s^2
>
> s^2 = 0.1 x 0.01 / 250
>
> s^2 = 0.001 / 250
>
> s = 0.002 seconds
Well, using the mass of the metal moved really has little to do with the
hammer strike. The energy is not used to move the metal. That is, knowing
how far it moved doesn't tell us the energy needed to move it. The energy
is used to accelerate the metal, and then, once accelerated, it moves
without the help of any energy. To stop the motion, the energy must then
be taken out of the metal and put somewhere else (into heat in the metal in
this case).
It takes very little energy to move 100 g of metal 1 cm no matter how it's
done, and that energy I suspect has very little to do with the much larger
forces and energy at work in the hammer strike.
The energy lost in the hammer strike ends up as heat in the metal and only
a small amount is used to actually move the metal a little bit before it
becomes heat.
> a 5kg hammer, hitting at 10m/s second and rebounding at 5 m/s 0.002
> seconds later has an acceleration of 7500m/s^2 (15m/s / 0.002s)
Looks good, but that .002 second number seems to be the result of an
invalid calculation.
> and hits with a force of 5 x 7500m/s = 37500 N (or 3.75 Tonnes if you
> prefer)
Seems reasonable, but I think you messed up the calculation.
From another approach...
If you know the metal compressed 1 cm in the strike, and we make the
assumption the hammer is undergoing constant deceleration in that 1 cm
distance (probably not too far from true), we can calculate time and force
without knowing anything about the mass of the metal being moved. We can
just calculate how much force was needed to stop the hammer at the bottom
of that 1 cm move.
Energy is force times distance, and all the 500 J of energy must be taken
out in that 1 cm (half gets added back later in the bounce). Constant
deceleration means constant force. So we just use energy = force x
distance:
500 = f x .01
f = 500 / .01
f = 50,000 N or 5.6 US short tons of force.
And knowing the force we can back calculate time with f=ma...
F = m a
50,000 N = 5 kg x .02 m / s^2
s^2 = 5 * .02 / 50,000
s = sqrt(.000002) = .0014 seconds.
Which is about the same as the .002 seconds you calculated anyway even
though as far as I can tell, the formula you used wasn't valid.
> BUT -
> there's a LOT of assumptions there -
Yes, there sure are.
> the real answer is:
> it depends.
For sure. But it's fun some times playing with numbers. Not as much fun
as hitting hot steel however.
