calculation of impact velocity needed to rap open Darby cuffs

Calculation (skip down to 'my opinion' if you are not an engineer)

Assume a Darby handcuff bolt has mass m. Assume the bolt spring is initially uncompressed (the locked state). Assume the handcuff is moving at velocity v when the handcuff strikes a massive object and comes to an instantaneous stop. The momentum of the bolt then moves the bolt forward (in the x direction, which unlocks the cuff) with initial velocity v, gradually getting slower as it is retarded by the force of the increasing spring compression x. The retarding force on the bolt at spring compression x is F = kx where k is the spring constant. The differential work done by the bolt movement against the spring compression is dW = Fdx = kxdx. So the total work needed at spring compression x is W = integral (kxdx) = k* integral (xdx) = k(D^2)/2 where D is the forward displacement of the bolt in the x direction, towards unlocking the cuff. Equating this work to the initial kinetic energy of the bolt ½ * m *v^2 we obtain an equation for the final displacement of the bolt D:
½* m*v^2 = k*(D^2)/2 or v^2 = (k/m)* D^2 = (k*D/m) * D
Now k*D is the final retarding force P exerted by the spring, when the bolt has moved distance D
Ie v^2 = (P/m)*D
Now the weight of the bolt in earths gravity is w = mg, so m=w/g and the final result for the rapping velocity v needed to move the bolt distance D is

v = square-root [ (P/w)*g *D ]

Conclusion:

the rapping velocity needed to open a darby is

square-root [ (P/w)*g *D ]

where P= pull force (in ounces) on the bolt needed to open the cuff, w= weight (in ounces) of the bolt, g=981 cm2/sec, D= distance bolt has to move (in cm) to open the cuff (the answer will will be in centimeters per second).

Numerical example:

If the bolt is 5mm diameter and is 5cm long, its volume is 1cc and its weight is 7.8gm = 0.27 oz. If the spring pull force P needed to open the cuff is 5lbs = 5*16=80 oz , and if the bolt has to move distance D= 0.5cm to open the cuff, the rapping velocity required is
square-root[(80/0.27)*981*0.5] = 380 cm/sec = 12.5 feet per second.

My opinion:

It should be practical to rap open darby cuffs which have a spring pull of about 5lbs or less with a good firm impact. This is roughly as my own experience indicates. Cuffs with pull forces as low as 5lbs have noticably weak springs. Most darby cuffs have spring pull forces above 5lbs, and personally I can't rap these open. Its also much harder to rap open darby leg irons (impossible?) because its very hard to hit hard enough whilst wearing them.

Hello, I will comment on your formula with and based upon my own experience rapping darbies,,,,,ok maybe not as much, Im white and cant rap
but having tried it over the years some work some dont and when really fresh springs are in the cuff its always led me to use the following formula

WTF!!! followed with some bruising

Calculation (skip down to 'my opinion' if you are not an engineer)

Assume a Darby handcuff bolt has mass m. Assume the bolt spring is initially uncompressed (the locked state). Assume the handcuff is moving at velocity v when the handcuff strikes a massive object and comes to an instantaneous stop. The momentum of the bolt then moves the bolt forward (in the x direction, which unlocks the cuff) with initial velocity v, gradually getting slower as it is retarded by the force of the increasing spring compression x. The retarding force on the bolt at spring compression x is F = kx where k is the spring constant. The differential work done by the bolt movement against the spring compression is dW = Fdx = kxdx. So the total work needed at spring compression x is W = integral (kxdx) = k* integral (xdx) = k(D^2)/2 where D is the forward displacement of the bolt in the x direction, towards unlocking the cuff. Equating this work to the initial kinetic energy of the bolt ½ * m *v^2 we obtain an equation for the final displacement of the bolt D:
½* m*v^2 = k*(D^2)/2 or v^2 = (k/m)* D^2 = (k*D/m) * D
Now k*D is the final retarding force P exerted by the spring, when the bolt has moved distance D
Ie v^2 = (P/m)*D
Now the weight of the bolt in earths gravity is w = mg, so m=w/g and the final result for the rapping velocity v needed to move the bolt distance D is

v = square-root [ (P/w)*g *D ]

Conclusion:

the rapping velocity needed to open a darby is

square-root [ (P/w)*g *D ]

where P= pull force (in ounces) on the bolt needed to open the cuff, w= weight (in ounces) of the bolt, g=981 cm2/sec, D= distance bolt has to move (in cm) to open the cuff (the answer will will be in centimeters per second).

Numerical example:

If the bolt is 5mm diameter and is 5cm long, its volume is 1cc and its weight is 7.8gm = 0.27 oz. If the spring pull force P needed to open the cuff is 5lbs = 5*16=80 oz , and if the bolt has to move distance D= 0.5cm to open the cuff, the rapping velocity required is
square-root[(80/0.27)*981*0.5] = 380 cm/sec = 12.5 feet per second.

My opinion:

It should be practical to rap open darby cuffs which have a spring pull of about 5lbs or less with a good firm impact. This is roughly as my own experience indicates. Cuffs with pull forces as low as 5lbs have noticably weak springs. Most darby cuffs have spring pull forces above 5lbs, and personally I can't rap these open. Its also much harder to rap open darby leg irons (impossible?) because its very hard to hit hard enough whilst wearing them.

Click to expand...

Dear Andrew,

That is impressive.

Here is how to do the superscript -- in this case, for something squared or taken to some other power -- so you don't need to suggest it by using the ^ character.

D² yields D².

Subscripts are similar. Here is the formula for water.

H₂O yields H₂O.

Cheers,

Jack
http://www.editing.org.uk/tanis/collectibles.htm

Calculation (skip down to 'my opinion' if you are not an engineer)

Assume a Darby handcuff bolt has mass m. Assume the bolt spring is initially uncompressed (the locked state). Assume the handcuff is moving at velocity v when the handcuff strikes a massive object and comes to an instantaneous stop. The momentum of the bolt then moves the bolt forward (in the x direction, which unlocks the cuff) with initial velocity v, gradually getting slower as it is retarded by the force of the increasing spring compression x. The retarding force on the bolt at spring compression x is F = kx where k is the spring constant. The differential work done by the bolt movement against the spring compression is dW = Fdx = kxdx. So the total work needed at spring compression x is W = integral (kxdx) = k* integral (xdx) = k(D^2)/2 where D is the forward displacement of the bolt in the x direction, towards unlocking the cuff. Equating this work to the initial kinetic energy of the bolt ½ * m *v^2 we obtain an equation for the final displacement of the bolt D:
½* m*v^2 = k*(D^2)/2 or v^2 = (k/m)* D^2 = (k*D/m) * D
Now k*D is the final retarding force P exerted by the spring, when the bolt has moved distance D
Ie v^2 = (P/m)*D
Now the weight of the bolt in earths gravity is w = mg, so m=w/g and the final result for the rapping velocity v needed to move the bolt distance D is

v = square-root [ (P/w)*g *D ]

Conclusion:

the rapping velocity needed to open a darby is

square-root [ (P/w)*g *D ]

where P= pull force (in ounces) on the bolt needed to open the cuff, w= weight (in ounces) of the bolt, g=981 cm2/sec, D= distance bolt has to move (in cm) to open the cuff (the answer will will be in centimeters per second).

Numerical example:

If the bolt is 5mm diameter and is 5cm long, its volume is 1cc and its weight is 7.8gm = 0.27 oz. If the spring pull force P needed to open the cuff is 5lbs = 5*16=80 oz , and if the bolt has to move distance D= 0.5cm to open the cuff, the rapping velocity required is
square-root[(80/0.27)*981*0.5] = 380 cm/sec = 12.5 feet per second.

My opinion:

It should be practical to rap open darby cuffs which have a spring pull of about 5lbs or less with a good firm impact. This is roughly as my own experience indicates. Cuffs with pull forces as low as 5lbs have noticably weak springs. Most darby cuffs have spring pull forces above 5lbs, and personally I can't rap these open. Its also much harder to rap open darby leg irons (impossible?) because its very hard to hit hard enough whilst wearing them.

Click to expand...

It didn't quite work out...

Don't separate anything with spaces

D 2 yields D2.

H 2 O yields H2O.



Cheers,

Jack
http://www.editing.org.uk/tanis/collectibles.htm

Calculation (skip down to 'my opinion' if you are not an engineer)

Assume a Darby handcuff bolt has mass m. Assume the bolt spring is initially uncompressed (the locked state). Assume the handcuff is moving at velocity v when the handcuff strikes a massive object and comes to an instantaneous stop. The momentum of the bolt then moves the bolt forward (in the x direction, which unlocks the cuff) with initial velocity v, gradually getting slower as it is retarded by the force of the increasing spring compression x. The retarding force on the bolt at spring compression x is F = kx where k is the spring constant. The differential work done by the bolt movement against the spring compression is dW = Fdx = kxdx. So the total work needed at spring compression x is W = integral (kxdx) = k* integral (xdx) = k(D^2)/2 where D is the forward displacement of the bolt in the x direction, towards unlocking the cuff. Equating this work to the initial kinetic energy of the bolt ½ * m *v^2 we obtain an equation for the final displacement of the bolt D:
½* m*v^2 = k*(D^2)/2 or v^2 = (k/m)* D^2 = (k*D/m) * D
Now k*D is the final retarding force P exerted by the spring, when the bolt has moved distance D
Ie v^2 = (P/m)*D
Now the weight of the bolt in earths gravity is w = mg, so m=w/g and the final result for the rapping velocity v needed to move the bolt distance D is

v = square-root [ (P/w)*g *D ]

Conclusion:

the rapping velocity needed to open a darby is

square-root [ (P/w)*g *D ]

where P= pull force (in ounces) on the bolt needed to open the cuff, w= weight (in ounces) of the bolt, g=981 cm2/sec, D= distance bolt has to move (in cm) to open the cuff (the answer will will be in centimeters per second).

Numerical example:

If the bolt is 5mm diameter and is 5cm long, its volume is 1cc and its weight is 7.8gm = 0.27 oz. If the spring pull force P needed to open the cuff is 5lbs = 5*16=80 oz , and if the bolt has to move distance D= 0.5cm to open the cuff, the rapping velocity required is
square-root[(80/0.27)*981*0.5] = 380 cm/sec = 12.5 feet per second.

My opinion:

It should be practical to rap open darby cuffs which have a spring pull of about 5lbs or less with a good firm impact. This is roughly as my own experience indicates. Cuffs with pull forces as low as 5lbs have noticably weak springs. Most darby cuffs have spring pull forces above 5lbs, and personally I can't rap these open. Its also much harder to rap open darby leg irons (impossible?) because its very hard to hit hard enough whilst wearing them.

Click to expand...

I'm very sorry. The Forum's HTML editor has defeated me.

Cheers,

Jack
http://www.editing.org.uk/tanis/collectibles.htm

Calculation (skip down to 'my opinion' if you are not an engineer)

Assume a Darby handcuff bolt has mass m. Assume the bolt spring is initially uncompressed (the locked state). Assume the handcuff is moving at velocity v when the handcuff strikes a massive object and comes to an instantaneous stop. The momentum of the bolt then moves the bolt forward (in the x direction, which unlocks the cuff) with initial velocity v, gradually getting slower as it is retarded by the force of the increasing spring compression x. The retarding force on the bolt at spring compression x is F = kx where k is the spring constant. The differential work done by the bolt movement against the spring compression is dW = Fdx = kxdx. So the total work needed at spring compression x is W = integral (kxdx) = k* integral (xdx) = k(D^2)/2 where D is the forward displacement of the bolt in the x direction, towards unlocking the cuff. Equating this work to the initial kinetic energy of the bolt ½ * m *v^2 we obtain an equation for the final displacement of the bolt D:
½* m*v^2 = k*(D^2)/2 or v^2 = (k/m)* D^2 = (k*D/m) * D
Now k*D is the final retarding force P exerted by the spring, when the bolt has moved distance D
Ie v^2 = (P/m)*D
Now the weight of the bolt in earths gravity is w = mg, so m=w/g and the final result for the rapping velocity v needed to move the bolt distance D is

v = square-root [ (P/w)*g *D ]

Conclusion:

the rapping velocity needed to open a darby is

square-root [ (P/w)*g *D ]

where P= pull force (in ounces) on the bolt needed to open the cuff, w= weight (in ounces) of the bolt, g=981 cm2/sec, D= distance bolt has to move (in cm) to open the cuff (the answer will will be in centimeters per second).

Numerical example:

If the bolt is 5mm diameter and is 5cm long, its volume is 1cc and its weight is 7.8gm = 0.27 oz. If the spring pull force P needed to open the cuff is 5lbs = 5*16=80 oz , and if the bolt has to move distance D= 0.5cm to open the cuff, the rapping velocity required is
square-root[(80/0.27)*981*0.5] = 380 cm/sec = 12.5 feet per second.

My opinion:

It should be practical to rap open darby cuffs which have a spring pull of about 5lbs or less with a good firm impact. This is roughly as my own experience indicates. Cuffs with pull forces as low as 5lbs have noticably weak springs. Most darby cuffs have spring pull forces above 5lbs, and personally I can't rap these open. Its also much harder to rap open darby leg irons (impossible?) because its very hard to hit hard enough whilst wearing them.

Click to expand...

Just write is as di-hydrogen monoxide...much easier!

Calculation (skip down to 'my opinion' if you are not an engineer)

Assume a Darby handcuff bolt has mass m. Assume the bolt spring is initially uncompressed (the locked state). Assume the handcuff is moving at velocity v when the handcuff strikes a massive object and comes to an instantaneous stop. The momentum of the bolt then moves the bolt forward (in the x direction, which unlocks the cuff) with initial velocity v, gradually getting slower as it is retarded by the force of the increasing spring compression x. The retarding force on the bolt at spring compression x is F = kx where k is the spring constant. The differential work done by the bolt movement against the spring compression is dW = Fdx = kxdx. So the total work needed at spring compression x is W = integral (kxdx) = k* integral (xdx) = k(D^2)/2 where D is the forward displacement of the bolt in the x direction, towards unlocking the cuff. Equating this work to the initial kinetic energy of the bolt ½ * m *v^2 we obtain an equation for the final displacement of the bolt D:
½* m*v^2 = k*(D^2)/2 or v^2 = (k/m)* D^2 = (k*D/m) * D
Now k*D is the final retarding force P exerted by the spring, when the bolt has moved distance D
Ie v^2 = (P/m)*D
Now the weight of the bolt in earths gravity is w = mg, so m=w/g and the final result for the rapping velocity v needed to move the bolt distance D is

v = square-root [ (P/w)*g *D ]

Conclusion:

the rapping velocity needed to open a darby is

square-root [ (P/w)*g *D ]

where P= pull force (in ounces) on the bolt needed to open the cuff, w= weight (in ounces) of the bolt, g=981 cm2/sec, D= distance bolt has to move (in cm) to open the cuff (the answer will will be in centimeters per second).

Numerical example:

If the bolt is 5mm diameter and is 5cm long, its volume is 1cc and its weight is 7.8gm = 0.27 oz. If the spring pull force P needed to open the cuff is 5lbs = 5*16=80 oz , and if the bolt has to move distance D= 0.5cm to open the cuff, the rapping velocity required is
square-root[(80/0.27)*981*0.5] = 380 cm/sec = 12.5 feet per second.

My opinion:

It should be practical to rap open darby cuffs which have a spring pull of about 5lbs or less with a good firm impact. This is roughly as my own experience indicates. Cuffs with pull forces as low as 5lbs have noticably weak springs. Most darby cuffs have spring pull forces above 5lbs, and personally I can't rap these open. Its also much harder to rap open darby leg irons (impossible?) because its very hard to hit hard enough whilst wearing them.

Click to expand...

If anyone wants a rappable pair of handcuffs, I have a pair that I'm willing to part with. I got them on ebay a few years ago, advertised as standard cuffs. But I have recently discovered that both sides can be rapped open with a firm impact. It does hurt the wrists a little to do this, but no bruising. The key also works normally. The secret is that the springs are both lighter than usual. I don't see any signs of modification, so I assume they were manufactured this way. Only markings I see are 'British made" on one bow. If interested, email me any offers or questions.

Calculation (skip down to 'my opinion' if you are not an engineer)

Assume a Darby handcuff bolt has mass m. Assume the bolt spring is initially uncompressed (the locked state). Assume the handcuff is moving at velocity v when the handcuff strikes a massive object and comes to an instantaneous stop. The momentum of the bolt then moves the bolt forward (in the x direction, which unlocks the cuff) with initial velocity v, gradually getting slower as it is retarded by the force of the increasing spring compression x. The retarding force on the bolt at spring compression x is F = kx where k is the spring constant. The differential work done by the bolt movement against the spring compression is dW = Fdx = kxdx. So the total work needed at spring compression x is W = integral (kxdx) = k* integral (xdx) = k(D^2)/2 where D is the forward displacement of the bolt in the x direction, towards unlocking the cuff. Equating this work to the initial kinetic energy of the bolt ½ * m *v^2 we obtain an equation for the final displacement of the bolt D:
½* m*v^2 = k*(D^2)/2 or v^2 = (k/m)* D^2 = (k*D/m) * D
Now k*D is the final retarding force P exerted by the spring, when the bolt has moved distance D
Ie v^2 = (P/m)*D
Now the weight of the bolt in earths gravity is w = mg, so m=w/g and the final result for the rapping velocity v needed to move the bolt distance D is

v = square-root [ (P/w)*g *D ]

Conclusion:

the rapping velocity needed to open a darby is

square-root [ (P/w)*g *D ]

where P= pull force (in ounces) on the bolt needed to open the cuff, w= weight (in ounces) of the bolt, g=981 cm2/sec, D= distance bolt has to move (in cm) to open the cuff (the answer will will be in centimeters per second).

Numerical example:

If the bolt is 5mm diameter and is 5cm long, its volume is 1cc and its weight is 7.8gm = 0.27 oz. If the spring pull force P needed to open the cuff is 5lbs = 5*16=80 oz , and if the bolt has to move distance D= 0.5cm to open the cuff, the rapping velocity required is
square-root[(80/0.27)*981*0.5] = 380 cm/sec = 12.5 feet per second.

My opinion:

It should be practical to rap open darby cuffs which have a spring pull of about 5lbs or less with a good firm impact. This is roughly as my own experience indicates. Cuffs with pull forces as low as 5lbs have noticably weak springs. Most darby cuffs have spring pull forces above 5lbs, and personally I can't rap these open. Its also much harder to rap open darby leg irons (impossible?) because its very hard to hit hard enough whilst wearing them.

Click to expand...

Hi, love it! we need more like you.

Ian McColl
(PS holding freely in my hand and striking several darbies firmly on an anvil from about 18 inches high, I was able to open one side of a pair , of 8 sets tried.)

Calculation (skip down to 'my opinion' if you are not an engineer)

Assume a Darby handcuff bolt has mass m. Assume the bolt spring is initially uncompressed (the locked state). Assume the handcuff is moving at velocity v when the handcuff strikes a massive object and comes to an instantaneous stop. The momentum of the bolt then moves the bolt forward (in the x direction, which unlocks the cuff) with initial velocity v, gradually getting slower as it is retarded by the force of the increasing spring compression x. The retarding force on the bolt at spring compression x is F = kx where k is the spring constant. The differential work done by the bolt movement against the spring compression is dW = Fdx = kxdx. So the total work needed at spring compression x is W = integral (kxdx) = k* integral (xdx) = k(D^2)/2 where D is the forward displacement of the bolt in the x direction, towards unlocking the cuff. Equating this work to the initial kinetic energy of the bolt ½ * m *v^2 we obtain an equation for the final displacement of the bolt D:
½* m*v^2 = k*(D^2)/2 or v^2 = (k/m)* D^2 = (k*D/m) * D
Now k*D is the final retarding force P exerted by the spring, when the bolt has moved distance D
Ie v^2 = (P/m)*D
Now the weight of the bolt in earths gravity is w = mg, so m=w/g and the final result for the rapping velocity v needed to move the bolt distance D is

v = square-root [ (P/w)*g *D ]

Conclusion:

the rapping velocity needed to open a darby is

square-root [ (P/w)*g *D ]

where P= pull force (in ounces) on the bolt needed to open the cuff, w= weight (in ounces) of the bolt, g=981 cm2/sec, D= distance bolt has to move (in cm) to open the cuff (the answer will will be in centimeters per second).

Numerical example:

If the bolt is 5mm diameter and is 5cm long, its volume is 1cc and its weight is 7.8gm = 0.27 oz. If the spring pull force P needed to open the cuff is 5lbs = 5*16=80 oz , and if the bolt has to move distance D= 0.5cm to open the cuff, the rapping velocity required is
square-root[(80/0.27)*981*0.5] = 380 cm/sec = 12.5 feet per second.

My opinion:

It should be practical to rap open darby cuffs which have a spring pull of about 5lbs or less with a good firm impact. This is roughly as my own experience indicates. Cuffs with pull forces as low as 5lbs have noticably weak springs. Most darby cuffs have spring pull forces above 5lbs, and personally I can't rap these open. Its also much harder to rap open darby leg irons (impossible?) because its very hard to hit hard enough whilst wearing them.

Click to expand...

Hello,
i have several darbys, most from Hiatt and one modern Hiatt/Thompson made from cast metal. The spring from the Hiatt/Thompson has nearly the same force than the other darbys, so i have tried to rap open the Hiatt/Thompson. I do not want to do this with the originals. I have rapped the cuff on a carpet to avoid scratches and it was possible to open bothcuffs several times. But i held the cuffs in my hands, they where not on my wrists. It is more difficult to do this with cuffs on the wrist, but possible, i have tried it with Peerless710 on my wrists for which you need nearly the same impact to open its doublelock by rapping.

The formulas are much easier: the spring in the cuff does not have force zero when cuff is closed, so the force is nearly constant when cuff is opened. Energy to open Cuff is E=F*d with E = energy, F = force of spring, d = distance lock bolt must move to open cuff. Also energy is E=v*v*m/2 with v = velocity of cuff when hitting floor, m = mass of locking bolt alone (plus a small fraction of spring).

So v*v*m/2 = F*d and
v = sqrt(2*F*d/m)

But it is much more realistic to make a real test, there are some factors which are not included by theory (e.g. friction)

Best Regards
Andy-58

Calculation (skip down to 'my opinion' if you are not an engineer)

Assume a Darby handcuff bolt has mass m. Assume the bolt spring is initially uncompressed (the locked state). Assume the handcuff is moving at velocity v when the handcuff strikes a massive object and comes to an instantaneous stop. The momentum of the bolt then moves the bolt forward (in the x direction, which unlocks the cuff) with initial velocity v, gradually getting slower as it is retarded by the force of the increasing spring compression x. The retarding force on the bolt at spring compression x is F = kx where k is the spring constant. The differential work done by the bolt movement against the spring compression is dW = Fdx = kxdx. So the total work needed at spring compression x is W = integral (kxdx) = k* integral (xdx) = k(D^2)/2 where D is the forward displacement of the bolt in the x direction, towards unlocking the cuff. Equating this work to the initial kinetic energy of the bolt ½ * m *v^2 we obtain an equation for the final displacement of the bolt D:
½* m*v^2 = k*(D^2)/2 or v^2 = (k/m)* D^2 = (k*D/m) * D
Now k*D is the final retarding force P exerted by the spring, when the bolt has moved distance D
Ie v^2 = (P/m)*D
Now the weight of the bolt in earths gravity is w = mg, so m=w/g and the final result for the rapping velocity v needed to move the bolt distance D is

v = square-root [ (P/w)*g *D ]

Conclusion:

the rapping velocity needed to open a darby is

square-root [ (P/w)*g *D ]

where P= pull force (in ounces) on the bolt needed to open the cuff, w= weight (in ounces) of the bolt, g=981 cm2/sec, D= distance bolt has to move (in cm) to open the cuff (the answer will will be in centimeters per second).

Numerical example:

If the bolt is 5mm diameter and is 5cm long, its volume is 1cc and its weight is 7.8gm = 0.27 oz. If the spring pull force P needed to open the cuff is 5lbs = 5*16=80 oz , and if the bolt has to move distance D= 0.5cm to open the cuff, the rapping velocity required is
square-root[(80/0.27)*981*0.5] = 380 cm/sec = 12.5 feet per second.

My opinion:

It should be practical to rap open darby cuffs which have a spring pull of about 5lbs or less with a good firm impact. This is roughly as my own experience indicates. Cuffs with pull forces as low as 5lbs have noticably weak springs. Most darby cuffs have spring pull forces above 5lbs, and personally I can't rap these open. Its also much harder to rap open darby leg irons (impossible?) because its very hard to hit hard enough whilst wearing them.

Click to expand...

From my experience, the old Thompsons(pre 20 cent.) seem to have a lighter spring than the Hiatts. I used to do a handcuff act where I pretended to trip and fall forward but really using the fall and angling the the cuffs to hit that sweet spot and pop open both at the same time. If you do that with Hiatts too many times, the "C" portion tends to deform as they used a different type of forged metal than Thompson that did not change shape.

Calculation (skip down to 'my opinion' if you are not an engineer)

Assume a Darby handcuff bolt has mass m. Assume the bolt spring is initially uncompressed (the locked state). Assume the handcuff is moving at velocity v when the handcuff strikes a massive object and comes to an instantaneous stop. The momentum of the bolt then moves the bolt forward (in the x direction, which unlocks the cuff) with initial velocity v, gradually getting slower as it is retarded by the force of the increasing spring compression x. The retarding force on the bolt at spring compression x is F = kx where k is the spring constant. The differential work done by the bolt movement against the spring compression is dW = Fdx = kxdx. So the total work needed at spring compression x is W = integral (kxdx) = k* integral (xdx) = k(D^2)/2 where D is the forward displacement of the bolt in the x direction, towards unlocking the cuff. Equating this work to the initial kinetic energy of the bolt ½ * m *v^2 we obtain an equation for the final displacement of the bolt D:
½* m*v^2 = k*(D^2)/2 or v^2 = (k/m)* D^2 = (k*D/m) * D
Now k*D is the final retarding force P exerted by the spring, when the bolt has moved distance D
Ie v^2 = (P/m)*D
Now the weight of the bolt in earths gravity is w = mg, so m=w/g and the final result for the rapping velocity v needed to move the bolt distance D is

v = square-root [ (P/w)*g *D ]

Conclusion:

the rapping velocity needed to open a darby is

square-root [ (P/w)*g *D ]

where P= pull force (in ounces) on the bolt needed to open the cuff, w= weight (in ounces) of the bolt, g=981 cm2/sec, D= distance bolt has to move (in cm) to open the cuff (the answer will will be in centimeters per second).

Numerical example:

If the bolt is 5mm diameter and is 5cm long, its volume is 1cc and its weight is 7.8gm = 0.27 oz. If the spring pull force P needed to open the cuff is 5lbs = 5*16=80 oz , and if the bolt has to move distance D= 0.5cm to open the cuff, the rapping velocity required is
square-root[(80/0.27)*981*0.5] = 380 cm/sec = 12.5 feet per second.

My opinion:

It should be practical to rap open darby cuffs which have a spring pull of about 5lbs or less with a good firm impact. This is roughly as my own experience indicates. Cuffs with pull forces as low as 5lbs have noticably weak springs. Most darby cuffs have spring pull forces above 5lbs, and personally I can't rap these open. Its also much harder to rap open darby leg irons (impossible?) because its very hard to hit hard enough whilst wearing them.

Click to expand...

RAPPUDY RAP RAP //// GIVE THEM A TAP //// THEY WILL OPEN //// JUST LIKE HE SPOKEN ////MY DARBYS ON //// MY DARBIE OFF TO YOU //// ANDREW !!!

LK D~D

Calculation (skip down to 'my opinion' if you are not an engineer)

Assume a Darby handcuff bolt has mass m. Assume the bolt spring is initially uncompressed (the locked state). Assume the handcuff is moving at velocity v when the handcuff strikes a massive object and comes to an instantaneous stop. The momentum of the bolt then moves the bolt forward (in the x direction, which unlocks the cuff) with initial velocity v, gradually getting slower as it is retarded by the force of the increasing spring compression x. The retarding force on the bolt at spring compression x is F = kx where k is the spring constant. The differential work done by the bolt movement against the spring compression is dW = Fdx = kxdx. So the total work needed at spring compression x is W = integral (kxdx) = k* integral (xdx) = k(D^2)/2 where D is the forward displacement of the bolt in the x direction, towards unlocking the cuff. Equating this work to the initial kinetic energy of the bolt ½ * m *v^2 we obtain an equation for the final displacement of the bolt D:
½* m*v^2 = k*(D^2)/2 or v^2 = (k/m)* D^2 = (k*D/m) * D
Now k*D is the final retarding force P exerted by the spring, when the bolt has moved distance D
Ie v^2 = (P/m)*D
Now the weight of the bolt in earths gravity is w = mg, so m=w/g and the final result for the rapping velocity v needed to move the bolt distance D is

v = square-root [ (P/w)*g *D ]

Conclusion:

the rapping velocity needed to open a darby is

square-root [ (P/w)*g *D ]

where P= pull force (in ounces) on the bolt needed to open the cuff, w= weight (in ounces) of the bolt, g=981 cm2/sec, D= distance bolt has to move (in cm) to open the cuff (the answer will will be in centimeters per second).

Numerical example:

If the bolt is 5mm diameter and is 5cm long, its volume is 1cc and its weight is 7.8gm = 0.27 oz. If the spring pull force P needed to open the cuff is 5lbs = 5*16=80 oz , and if the bolt has to move distance D= 0.5cm to open the cuff, the rapping velocity required is
square-root[(80/0.27)*981*0.5] = 380 cm/sec = 12.5 feet per second.

My opinion:

It should be practical to rap open darby cuffs which have a spring pull of about 5lbs or less with a good firm impact. This is roughly as my own experience indicates. Cuffs with pull forces as low as 5lbs have noticably weak springs. Most darby cuffs have spring pull forces above 5lbs, and personally I can't rap these open. Its also much harder to rap open darby leg irons (impossible?) because its very hard to hit hard enough whilst wearing them.

Click to expand...

Wow, a physics discussion on a handcuff collectors web site. WOW.

Now the bad news, as a Physics Teacher, I feel the need to point out that one of the most common misconceptions in Physics, is to ignore friction. On paper, this is often done to allow the students to focus on the topic at hand and not worry about real-world, hard to calculate side issues.

In the real world you cannot do that. The inner workings of a handcuff is a small compact tight space with small moving parts. There has to be some friction involved.

Sadly, the calculation of friction is not easy, and worse it changes with every other change. If the bolt moves faster or slower, the friction changes. This may be a strong source of error, or it may matter little, but as a Physics teacher I felt inclined to point it out.

Still I think this is a great topic, and I would love permission to re-bullish it in "The chain Letter" the monthly newsletter of the international escapologist society.

Thanks,

Cliff


check us out! www.tiesociety.info

Calculation (skip down to 'my opinion' if you are not an engineer)

Assume a Darby handcuff bolt has mass m. Assume the bolt spring is initially uncompressed (the locked state). Assume the handcuff is moving at velocity v when the handcuff strikes a massive object and comes to an instantaneous stop. The momentum of the bolt then moves the bolt forward (in the x direction, which unlocks the cuff) with initial velocity v, gradually getting slower as it is retarded by the force of the increasing spring compression x. The retarding force on the bolt at spring compression x is F = kx where k is the spring constant. The differential work done by the bolt movement against the spring compression is dW = Fdx = kxdx. So the total work needed at spring compression x is W = integral (kxdx) = k* integral (xdx) = k(D^2)/2 where D is the forward displacement of the bolt in the x direction, towards unlocking the cuff. Equating this work to the initial kinetic energy of the bolt ½ * m *v^2 we obtain an equation for the final displacement of the bolt D:
½* m*v^2 = k*(D^2)/2 or v^2 = (k/m)* D^2 = (k*D/m) * D
Now k*D is the final retarding force P exerted by the spring, when the bolt has moved distance D
Ie v^2 = (P/m)*D
Now the weight of the bolt in earths gravity is w = mg, so m=w/g and the final result for the rapping velocity v needed to move the bolt distance D is

v = square-root [ (P/w)*g *D ]

Conclusion:

the rapping velocity needed to open a darby is

square-root [ (P/w)*g *D ]

where P= pull force (in ounces) on the bolt needed to open the cuff, w= weight (in ounces) of the bolt, g=981 cm2/sec, D= distance bolt has to move (in cm) to open the cuff (the answer will will be in centimeters per second).

Numerical example:

If the bolt is 5mm diameter and is 5cm long, its volume is 1cc and its weight is 7.8gm = 0.27 oz. If the spring pull force P needed to open the cuff is 5lbs = 5*16=80 oz , and if the bolt has to move distance D= 0.5cm to open the cuff, the rapping velocity required is
square-root[(80/0.27)*981*0.5] = 380 cm/sec = 12.5 feet per second.

My opinion:

It should be practical to rap open darby cuffs which have a spring pull of about 5lbs or less with a good firm impact. This is roughly as my own experience indicates. Cuffs with pull forces as low as 5lbs have noticably weak springs. Most darby cuffs have spring pull forces above 5lbs, and personally I can't rap these open. Its also much harder to rap open darby leg irons (impossible?) because its very hard to hit hard enough whilst wearing them.

Click to expand...

Make that "re-publish"

Calculation (skip down to 'my opinion' if you are not an engineer)

Assume a Darby handcuff bolt has mass m. Assume the bolt spring is initially uncompressed (the locked state). Assume the handcuff is moving at velocity v when the handcuff strikes a massive object and comes to an instantaneous stop. The momentum of the bolt then moves the bolt forward (in the x direction, which unlocks the cuff) with initial velocity v, gradually getting slower as it is retarded by the force of the increasing spring compression x. The retarding force on the bolt at spring compression x is F = kx where k is the spring constant. The differential work done by the bolt movement against the spring compression is dW = Fdx = kxdx. So the total work needed at spring compression x is W = integral (kxdx) = k* integral (xdx) = k(D^2)/2 where D is the forward displacement of the bolt in the x direction, towards unlocking the cuff. Equating this work to the initial kinetic energy of the bolt ½ * m *v^2 we obtain an equation for the final displacement of the bolt D:
½* m*v^2 = k*(D^2)/2 or v^2 = (k/m)* D^2 = (k*D/m) * D
Now k*D is the final retarding force P exerted by the spring, when the bolt has moved distance D
Ie v^2 = (P/m)*D
Now the weight of the bolt in earths gravity is w = mg, so m=w/g and the final result for the rapping velocity v needed to move the bolt distance D is

v = square-root [ (P/w)*g *D ]

Conclusion:

the rapping velocity needed to open a darby is

square-root [ (P/w)*g *D ]

where P= pull force (in ounces) on the bolt needed to open the cuff, w= weight (in ounces) of the bolt, g=981 cm2/sec, D= distance bolt has to move (in cm) to open the cuff (the answer will will be in centimeters per second).

Numerical example:

If the bolt is 5mm diameter and is 5cm long, its volume is 1cc and its weight is 7.8gm = 0.27 oz. If the spring pull force P needed to open the cuff is 5lbs = 5*16=80 oz , and if the bolt has to move distance D= 0.5cm to open the cuff, the rapping velocity required is
square-root[(80/0.27)*981*0.5] = 380 cm/sec = 12.5 feet per second.

My opinion:

It should be practical to rap open darby cuffs which have a spring pull of about 5lbs or less with a good firm impact. This is roughly as my own experience indicates. Cuffs with pull forces as low as 5lbs have noticably weak springs. Most darby cuffs have spring pull forces above 5lbs, and personally I can't rap these open. Its also much harder to rap open darby leg irons (impossible?) because its very hard to hit hard enough whilst wearing them.

Click to expand...

Cliff,
I have played with quite a number of darbies and know the tolerances on the inside. Friction has very little to do since it is almost floating when they impact a hard surface. I say this because there is so much play in the locks and what little there is so negligible to have much impact on the calculations. What you could expound on is the optimum point of impact and angle. The relationship of the lock tube to the hinge plays a crucial role. The mass being off to one side of the hinge naturally wants to travel outward and when the bolt retracts far enough, it flies right open.

-Dorson

Calculation (skip down to 'my opinion' if you are not an engineer)

Assume a Darby handcuff bolt has mass m. Assume the bolt spring is initially uncompressed (the locked state). Assume the handcuff is moving at velocity v when the handcuff strikes a massive object and comes to an instantaneous stop. The momentum of the bolt then moves the bolt forward (in the x direction, which unlocks the cuff) with initial velocity v, gradually getting slower as it is retarded by the force of the increasing spring compression x. The retarding force on the bolt at spring compression x is F = kx where k is the spring constant. The differential work done by the bolt movement against the spring compression is dW = Fdx = kxdx. So the total work needed at spring compression x is W = integral (kxdx) = k* integral (xdx) = k(D^2)/2 where D is the forward displacement of the bolt in the x direction, towards unlocking the cuff. Equating this work to the initial kinetic energy of the bolt ½ * m *v^2 we obtain an equation for the final displacement of the bolt D:
½* m*v^2 = k*(D^2)/2 or v^2 = (k/m)* D^2 = (k*D/m) * D
Now k*D is the final retarding force P exerted by the spring, when the bolt has moved distance D
Ie v^2 = (P/m)*D
Now the weight of the bolt in earths gravity is w = mg, so m=w/g and the final result for the rapping velocity v needed to move the bolt distance D is

v = square-root [ (P/w)*g *D ]

Conclusion:

the rapping velocity needed to open a darby is

square-root [ (P/w)*g *D ]

where P= pull force (in ounces) on the bolt needed to open the cuff, w= weight (in ounces) of the bolt, g=981 cm2/sec, D= distance bolt has to move (in cm) to open the cuff (the answer will will be in centimeters per second).

Numerical example:

If the bolt is 5mm diameter and is 5cm long, its volume is 1cc and its weight is 7.8gm = 0.27 oz. If the spring pull force P needed to open the cuff is 5lbs = 5*16=80 oz , and if the bolt has to move distance D= 0.5cm to open the cuff, the rapping velocity required is
square-root[(80/0.27)*981*0.5] = 380 cm/sec = 12.5 feet per second.

My opinion:

It should be practical to rap open darby cuffs which have a spring pull of about 5lbs or less with a good firm impact. This is roughly as my own experience indicates. Cuffs with pull forces as low as 5lbs have noticably weak springs. Most darby cuffs have spring pull forces above 5lbs, and personally I can't rap these open. Its also much harder to rap open darby leg irons (impossible?) because its very hard to hit hard enough whilst wearing them.

Click to expand...

Dorson, I bow to your experience with the devices. Of course uneven mass lends itself to Torque, and that is a whole nother story. Yes, the angel of attack would be very important. On that we agree.

Calculation (skip down to 'my opinion' if you are not an engineer)

Assume a Darby handcuff bolt has mass m. Assume the bolt spring is initially uncompressed (the locked state). Assume the handcuff is moving at velocity v when the handcuff strikes a massive object and comes to an instantaneous stop. The momentum of the bolt then moves the bolt forward (in the x direction, which unlocks the cuff) with initial velocity v, gradually getting slower as it is retarded by the force of the increasing spring compression x. The retarding force on the bolt at spring compression x is F = kx where k is the spring constant. The differential work done by the bolt movement against the spring compression is dW = Fdx = kxdx. So the total work needed at spring compression x is W = integral (kxdx) = k* integral (xdx) = k(D^2)/2 where D is the forward displacement of the bolt in the x direction, towards unlocking the cuff. Equating this work to the initial kinetic energy of the bolt ½ * m *v^2 we obtain an equation for the final displacement of the bolt D:
½* m*v^2 = k*(D^2)/2 or v^2 = (k/m)* D^2 = (k*D/m) * D
Now k*D is the final retarding force P exerted by the spring, when the bolt has moved distance D
Ie v^2 = (P/m)*D
Now the weight of the bolt in earths gravity is w = mg, so m=w/g and the final result for the rapping velocity v needed to move the bolt distance D is

v = square-root [ (P/w)*g *D ]

Conclusion:

the rapping velocity needed to open a darby is

square-root [ (P/w)*g *D ]

where P= pull force (in ounces) on the bolt needed to open the cuff, w= weight (in ounces) of the bolt, g=981 cm2/sec, D= distance bolt has to move (in cm) to open the cuff (the answer will will be in centimeters per second).

Numerical example:

If the bolt is 5mm diameter and is 5cm long, its volume is 1cc and its weight is 7.8gm = 0.27 oz. If the spring pull force P needed to open the cuff is 5lbs = 5*16=80 oz , and if the bolt has to move distance D= 0.5cm to open the cuff, the rapping velocity required is
square-root[(80/0.27)*981*0.5] = 380 cm/sec = 12.5 feet per second.

My opinion:

It should be practical to rap open darby cuffs which have a spring pull of about 5lbs or less with a good firm impact. This is roughly as my own experience indicates. Cuffs with pull forces as low as 5lbs have noticably weak springs. Most darby cuffs have spring pull forces above 5lbs, and personally I can't rap these open. Its also much harder to rap open darby leg irons (impossible?) because its very hard to hit hard enough whilst wearing them.

Click to expand...

"Yes, the angel of attack would be very important."



Angel of Attack