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From: Sam Wormley on 16 Apr 2007 11:06
GregS wrote:
> In article <86LUh.58733$oV.58014(a)attbi_s21>, Sam Wormley <swormley1(a)mchsi.com> wrote:
>> GregS wrote:
>>> I need some assistance to figure out a problem. i was looking for an online
>> calculator.
>>> I have a weigth mass, and I want to drop it at different heights, and compute
>>> the impact force. The impact area would also affect the actual force.
>>> Its about a 1 lb weight and a drop of 4 to 10 inches. I was planning on
>> buying
>>> a load cell of 100 lbs capacity to try to measure this, and I wanted to know
>> if this
>>> was enough.
>>>
>>> greg
>>
>> Try: http://hyperphysics.phy-astr.gsu.edu/hbase/impulse.html
>
> I have distance, but velocity is ????
>
> greg

http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#ffall
v(h) = sqrt(2*g*h)
From: John C. Polasek on 16 Apr 2007 11:34
On Mon, 16 Apr 2007 13:36:03 GMT, szekeres(a)pitt.edu (GregS) wrote:

>I need some assistance to figure out a problem. i was looking for an online calculator.
>I have a weigth mass, and I want to drop it at different heights, and compute
>the impact force. The impact area would also affect the actual force.
>Its about a 1 lb weight and a drop of 4 to 10 inches. I was planning on buying
>a load cell of 100 lbs capacity to try to measure this, and I wanted to know if this
>was enough.
>
>greg
There is no defined impact force. It entirely depends on how long it
takes to stop. If it drops into something soft, the force can be quite
low.
You can do some arithmetic as follows first to find the velocity:
V = sqrt(2g*h)
is the drop velocity falling h ft. where g is 32 ft/ss so
Vfps = 8*sqrt(hft)
Example for 6 inches or 1/2 foot V = 8*.707 =5.656 ft/sec.
A 1 pound weight is 1/32 slug or .03125 slugs
If the force waveform is a square (it isn't) then
MV = FT
M is pounds/32 slugs. T is the stopping time. F is the constant force
over that time.
So F = MV/T
You should have a load cell into an oscilloscope and try to record the
wave form. The area under the curve is integral FdT. The amplitude
will be the force.
John Polasek

From: JWS on 16 Apr 2007 12:41
Extending the previous message a bit:

JWS wrote:

> The formula is
>
> F_dynamic t_fall
> --------- = 1 + -------
> F_static t_sound
> This assumes that your load cell is infinitely stiff, i.e., it
> does not act as some kind of cushion which absorbs energy from the
> fall, thereby reducing the impact force. I think that measuring
> impact forces is quite a challenge for experimenters. I'd be very
> interested in your results. See http://www.jw-stumpel.nl/bounce.html

Please do not buy a load cell yet.

Let's assume that

-- your 1 lb weight is made of iron
-- it is fashioned into a cylinder of 2 cm diameter
-- by "1 lb" you mean 500 grams (0.5 kg).
-- at your location, g is 10 m/s².
-- the cylinder falls vertically from a height (= distance from
bottom of cylinder to impact point) of 20 cm (about 8 inches).

According to Wikipedia, the specific mass of iron is 7.86 g/cm^3,
and the speed of sound in iron is 5120 m/sec.

Then the height, L, of the iron cylinder will be 20.25 cm. (BTW
dropping it completely vertically will be a challenge. I suggest
using a vacuum "suction" system; open the tap when you want to
drop your sample).

The impact time (duration of the impact) will be 2L/v = 0.405/5120
= about 80 microseconds. This, I think, will be your major
challenge, because in order to measure anything at all, the
response time of your load cell must be far below this; let's say
5 microseconds or less, to get a reasonable number of measurement
points on the oscilloscope. AFAIK most load cells are designed for
measuring quasi-static loads, with relatively slow response times.

The impact force will be (according to the formula I gave in the
previous message) 5*(1 + sqrt(0.4/10)/(0.2025/5120))= 25289
newtons, or about 5060 pounds of force. A load cell of 100 lb
capacity will be vastly outclassed. Impact forces are REALLY big
(as the first cavemen discovered, when they attacked bears by
hitting them with stones).

So by all means order your load cell, but specify:

-- dynamic range: in excess of 25000 Newtons (minimum)
-- response time: less than 4 microseconds.

I do not know if such things exist. Piezo-electric elements?

JWS

From: GregS on 16 Apr 2007 12:52
In article <4623a74e$0$3748$ba620dc5(a)text.nova.planet.nl>, JWS <jws(a)my.home> wrote:
>Extending the previous message a bit:
>
>JWS wrote:
>
>> The formula is
>>
>> F_dynamic t_fall
>> --------- = 1 + -------
>> F_static t_sound
>> This assumes that your load cell is infinitely stiff, i.e., it
>> does not act as some kind of cushion which absorbs energy from the
>> fall, thereby reducing the impact force. I think that measuring
>> impact forces is quite a challenge for experimenters. I'd be very
>> interested in your results. See http://www.jw-stumpel.nl/bounce.html
>
>Please do not buy a load cell yet.
>
>Let's assume that
>
>-- your 1 lb weight is made of iron
>-- it is fashioned into a cylinder of 2 cm diameter
>-- by "1 lb" you mean 500 grams (0.5 kg).
>-- at your location, g is 10 m/s².
>-- the cylinder falls vertically from a height (= distance from
> bottom of cylinder to impact point) of 20 cm (about 8 inches).
>
>According to Wikipedia, the specific mass of iron is 7.86 g/cm^3,
>and the speed of sound in iron is 5120 m/sec.
>
>Then the height, L, of the iron cylinder will be 20.25 cm. (BTW
>dropping it completely vertically will be a challenge. I suggest
>using a vacuum "suction" system; open the tap when you want to
>drop your sample).
>
>The impact time (duration of the impact) will be 2L/v = 0.405/5120
>= about 80 microseconds. This, I think, will be your major
>challenge, because in order to measure anything at all, the
>response time of your load cell must be far below this; let's say
>5 microseconds or less, to get a reasonable number of measurement
>points on the oscilloscope. AFAIK most load cells are designed for
>measuring quasi-static loads, with relatively slow response times.
>
>The impact force will be (according to the formula I gave in the
>previous message) 5*(1 + sqrt(0.4/10)/(0.2025/5120))= 25289
>newtons, or about 5060 pounds of force. A load cell of 100 lb
>capacity will be vastly outclassed. Impact forces are REALLY big
>(as the first cavemen discovered, when they attacked bears by
>hitting them with stones).
>
>So by all means order your load cell, but specify:
>
> -- dynamic range: in excess of 25000 Newtons (minimum)
> -- response time: less than 4 microseconds.
>
>I do not know if such things exist. Piezo-electric elements?

Thanks much. Well, by the end of this I will be a lot smarter.
Trying to document a gillotine force. The actual impact area will be much
smaller than the actual weight. If I can manage the formulas, perhaps
measurment is unecessary, except for the actual static weight of
the brass cylinder.
My intended transducer.
http://www.meas-spec.com/mymsi/download/pdf/english/sensors/LoadCellFC2122.pdf


greg
From: JWS on 16 Apr 2007 13:54
GregS wrote:
> In article <4623a74e$0$3748$ba620dc5(a)text.nova.planet.nl>, JWS
> <jws(a)my.home> wrote:

>> So by all means order your load cell, but specify:
>>
>> -- dynamic range: in excess of 25000 Newtons (minimum) --
>> response time: less than 4 microseconds.
>>
>> I do not know if such things exist. Piezo-electric elements?

> Thanks much. Well, by the end of this I will be a lot smarter.
> Trying to document a gillotine force. The actual impact area
> will be much smaller than the actual weight. If I can manage
> the formulas, perhaps measurment is unecessary, except for the
> actual static weight of the brass cylinder.

ehh.. which brass cylinder?

> My intended transducer.
> http://www.meas-spec.com/mymsi/download/pdf/english/sensors/LoadCellFC2122.pdf

This transducer says it has "fast response time", but it does not
say how fast. I doubt it would be as fast as a few microseconds.
And the maximum dynamic range is 100 lbs of force (500 Newtons).
For measuring "pure" impact forces (e.g. for investigating the
9/11 building collapses) this would be quite inadequate.

But-- when you are really talking about the force of a guillotine
(I assume, the force exercised by this instrument upon a victim's
neck) the game is totally different. There we are no longer
talking about an "infinitely-unyielding ground", but rather about
something like "work of separation", "work of fracture", etc.; in
other words, about "energy" rather than "force". I am sure some
experts on this list can throw more light on this -- but I cannot
at this moment.
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