How to calculate the impact of a police K9

[DougCoMo]

Hi everyone, I'm a police dog trainer and am trying to explain to people at demos the impact of a police dog on a helper wearing a body bite suit at impact. So, a 75 pound German Shepherd traveling 28 mph for 60 yards impacting a 185 pound person in a body bite suit. I'm trying to determine the change in pounds of energy. The calculations in joules/Newtons I'm seeing on online calculators is abstract for people not familiar with the terms. Any input with the formula is greatly appreciated! Thx

 

[PeroK]

Summary:: What is the calculation for a police K9 impacting a body bite suit at speed

Hi everyone, I'm a police dog trainer and am trying to explain to people at demos the impact of a police dog on a helper wearing a body bite suit at impact. So, a 75 pound German Shepherd traveling 28 mph for 60 yards impacting a 185 pound person in a body bite suit. I'm trying to determine the change in pounds of energy. The calculations in joules/Newtons I'm seeing on online calculators is abstract for people not familiar with the terms. Any input with the formula is greatly appreciated! Thx

What are you actually trying to calculate? Whether someone gets knocked over, suffers bruising, breaks a bone?

In general, an impact depends on many factors. If someone is off-balance, then they'll be knocked over. If they are braced, then there may be a much larger force of impact.

 

[jbriggs444]

I'm trying to determine the change in pounds of energy.

Pounds are not units of energy. Nor is force the only figure of merit for an impact.

For an armored target, "impulse" (i.e. momentum) is a good figure of merit. It is not measured in pounds either.

And no, you can't convert it to pounds. This whole idea is fundamentally misguided. The notion that you can find a big number and a familiar unit to give people a correct understanding is horse hockey.

 

[DougCoMo]

The point I'm trying to reach is if the helper in the suit doesn't move at all or only perhaps 18 inches upon impact, how much impact occurs numerically (the increase in lbs energy?). The dogs can break canines, have disc injuries etc. We're trying to relate that the weight at speed needs to be well absorbed through movement (less direct impact) and then returning the dog quickly back to the ground. I had someone tell me it's like being impacted by 566 pounds and I have no idea if this even close.

 

[DougCoMo]

Pounds are not units of energy. Nor is force the only figure of merit for an impact.

For an armored target, "impulse" (i.e. momentum) is a good figure of merit. It is not measured in pounds either.

And no, you can't convert it to pounds. This whole idea is fundamentally misguided. The notion that you can find a big number and a familiar unit to give people a correct understanding is horse hockey.

Super, that's what I'm here to find out exactly. What is the best way to express this to the lay person?

 

[jbriggs444]

See #2 by @PeroK.

What are you trying to convey?

 

[DougCoMo]

What are you actually trying to calculate? Whether someone gets knocked over, suffers bruising, breaks a bone?

In general, an impact depends on many factors. If someone is off-balance, then they'll be knocked over. If they are braced, then there may be a much larger force of impact.

I'm trying to convey, in terms a layperson can understand, 1. the impact that is created by a 75 pound dog traveling 28 mph for 60m into a helper wearing a body bite suit (185 pound person). And 2. the difference when that a 185 pound person moves only 18 inches on impact. Thanks in advance.

 

[jbriggs444]

I'm trying to convey, in terms a layperson can understand, 1. the impact that is created by a 75 pound dog traveling 28 mph for 60m into a helper wearing a body bite suit (185 pound person). And 2. the difference when that a 185 pound person moves only 18 inches on impact. Thanks in advance.

"The impact".

Unfortunately, you have not told us what it is about the impact that you want to quantify.

"The difference".

The difference between what and what else?Back of the envelope...

So we have a 75 pound dog arriving at 28 miles per hour. We have an armored target that masses perhaps 200 pounds.

This amounts to an inelastic collision. We can apply a conservation of momentum argument to calculate that the man+dog tangle will have a resulting speed of $\frac{28 \times 75}{75+200}=8$ miles per hour.

If the man is able to lean forward at, perhaps, a 15 or 20 degree angle (1/4 to 1/3 of a gee) then 8 miles per hour (12 feet per second) will take almost one second to cancel. That'll knock a man back about six feet.

If you want to get that down to 18 inches, you'll need to be resisting with a full gee for about 1/3 of a second with a lean angle of 45 degrees. One could cheat this down a bit by moving forward to meet the dog.

The initial impact will involve forces significantly greater than this, albeit for a shorter duration as the dog decelerates to match speeds with the man.

 

[PeroK]

I'm trying to convey, in terms a layperson can understand, 1. the impact that is created by a 75 pound dog traveling 28 mph for 60m into a helper wearing a body bite suit (185 pound person). And 2. the difference when that a 185 pound person moves only 18 inches on impact. Thanks in advance.

I suspect you won't find the answer to this here because this would require specific research. The equations of physics help, of course, but calculations also require data. The size, shape and hardness of the dog are critical. Imagine a 75 pound rock instead of a dog. The difference is the hardness of the object that hits you.

At some point in your analysis you are going to have to specify the physical characteristics of the dog (as opposed to a rock or anything else). That's vital data.

Also, there's clearly an element of randomness to this. If a child falls off a bike, one day they get up unharmed and the next day they break their wrist, or worse. There cannot be any equation to say what happens when a child falls off a bike. The extent of the injury is random and can only be assessed by detailed specific research.

 

[DrStupid]

So, a 75 pound German Shepherd traveling 28 mph for 60 yards impacting a 185 pound person in a body bite suit.

If the person is off-balance it will be pushed back with about 8 mph. Maybe that gives an impression of the impact.

And 2. the difference when that a 185 pound person moves only 18 inches on impact.

You can calculate the corresponding force (in horizontal direction) by

$F = \frac{{m_{dog} \cdot v_{dog}^2 }}{{2 \cdot displacement}}$

 

[DougCoMo]

Thanks everyone! Here are some numbers I was given and I'll type this exactly as it was given to me. From what I'm reading here, this calculation is erroneous.

"Here is the exact calculation based on lbs of force. A K9 that weighs 75lbs traveling 27 mph upon impact of a decoy who only travels 18 inches upon impact... Force at point of impact: 566 lbs of force."

I'm told this was calculated by a junior college physics teacher. No clue how it was arrived at... does this make sense to anyone here?

 

[PeroK]

Thanks everyone! Here are some numbers I was given and I'll type this exactly as it was given to me. From what I'm reading here, this calculation is erroneous.

"Here is the exact calculation based on lbs of force. A K9 that weighs 75lbs traveling 27 mph upon impact of a decoy who only travels 18 inches upon impact... Force at point of impact: 566 lbs of force."

I'm told this was calculated by a junior college physics teacher. No clue how it was arrived at... does this make sense to anyone here?

It's easy to come up with a number like that, but what does it mean? You could go back to the physics teacher and ask what if it's a 75 lb rock? Why wouldn't you get the same answer?

At what point in the calculation does the fact that it's a dog and not a rock become relevant?

 

[jbriggs444]

Thanks everyone! Here are some numbers I was given and I'll type this exactly as it was given to me. From what I'm reading here, this calculation is erroneous.

"Here is the exact calculation based on lbs of force. A K9 that weighs 75lbs traveling 27 mph upon impact of a decoy who only travels 18 inches upon impact... Force at point of impact: 566 lbs of force."

I'm told this was calculated by a junior college physics teacher. No clue how it was arrived at... does this make sense to anyone here?

Let's try to reverse engineer the answer.

566 pounds of force acting on a 75 pound dog over an 18 inch displacement.

The resulting acceleration is 566/75 = 7.5 gees. 241 feet per second per second.

We need to get a 1.5 foot displacement with an acceleration of 241 feet per second². How much time will that take?
$$s=\frac{1}{2}at^2$$ $$t=\sqrt{\frac{2s}{a}}$$If we work that out, it's $\sqrt{\frac{3}{241}}$=1/9 of a second.

In 1/9 of a second, a 241 feet per second acceleration can get us to 27 feet per second.

The dog is actually moving at 27 miles per hour. It sounds like the junior college physics teacher forgot to convert miles per hour to feet per second.

Nonetheless, it is a decent calculation of the force required to decelerate the dog from 27 feet per second to zero over a distance moved of 18 inches. It does rely on the simplifying assumption that acceleration is constant. And it ignores the man entirely.

A more proper analysis might try to break the problem down and analyze the interaction of dog with man (as the dog is brought to a stop relative to the man) and the interaction of man with ground (as the man brings himself together with the dog to a stop). It is the latter interaction that I understand to cover 18 inches. A key problem is that we are left with little information to characterize the impact of dog on man -- no 18 inches to go on. But 18 inches is still a decent guess -- it is approximately the dog's radius.

 

[DrStupid]

"Here is the exact calculation based on lbs of force. A K9 that weighs 75lbs traveling 27 mph upon impact of a decoy who only travels 18 inches upon impact... Force at point of impact: 566 lbs of force."

That means in reasonable units

$m_{dog} = 34kg$
$v_{dog} = 12{\textstyle{m \over s}}$
$displacement = 0.457m$

and gives an average horizontal force of

$F = \frac{{m_{dog} \cdot v_{dog}^2 }}{{2 \cdot displacement}} = 5360 N$

(corresponding to 1200 lbs) acting between the dog and the decoy until it stops. This calculation with a different approach (based on kinetic energy and work) confirms jbriggs444’s result above. The junior college physics teacher most probably confused the units for the speed.

 

[Lnewqban]

The point I'm trying to reach is if the helper in the suit doesn't move at all or only perhaps 18 inches upon impact, how much impact occurs numerically (the increase in lbs energy?). The dogs can break canines, have disc injuries etc. We're trying to relate that the weight at speed needs to be well absorbed through movement (less direct impact) and then returning the dog quickly back to the ground. I had someone tell me it's like being impacted by 566 pounds and I have no idea if this even close.

It seems to me that a direct impact between the centers of mass of running-jumping dog and static suspect seldom occur.
By watching these slow motion videos, we can see that the dog goes for one of the arms of the subject, which makes it use some of its impulse to jump in order to bite higher.
Then, a natural rotation of the suspect's body happens, which dramatically reduces the efect of the impact on the balance of the subject.

The fun part starts when the impulse of the dog takes it to the opposite side of the original trajectory and pulls the entire body of the suspect via its canines bitting his arm.
I believe that it is at that point when the dogs can break canines and have disc injuries, because the material of the protecting gear does not tear and yield as flesh and skin do.
The training subject does not feel pain either, therefore he does not go along with the impulse of the body of the dog, like a real suspect would do naturally trying to minimize pain and flesh damage.

In nature, the canines are strong if they can penetrate skin and flesh of an animal, because as the wolf shakes his head and body to tear flesh and to increase pain and bleeding, the associate huge forces are uniformly distribuited closer to the root of the canines.
Bitting on the surface of the protective gear makes the canines of a police dog more vulnerable because the forces that could break them are applied to the tip rather than the section closer to the gum (greater moment or torque applied at the root).
Same for the spine discs, if the spine is no aligned when traction or compression or twisting forces happen.

Perhaps adding some elasticity or yielding capability to the usual areas of bites in the garmet could aliviate dog's injuries, as well as training on more yielding movements to "pain" from the helpers.

All the above makes any calculation complicated.
A way to experiment about the effect of pure impact could be launching a bag of sand against the body of a standing man.
That man could try catching that bag off center and try holding it while it rotates to the opposite side.
Of course, that would leave the pain effect out of the situation.

 

[DaveE]

Summary:: What is the calculation for a police K9 impacting a body bite suit at speed

Hi everyone, I'm a police dog trainer and am trying to explain to people at demos the impact of a police dog on a helper wearing a body bite suit at impact. So, a 75 pound German Shepherd traveling 28 mph for 60 yards impacting a 185 pound person in a body bite suit. I'm trying to determine the change in pounds of energy. The calculations in joules/Newtons I'm seeing on online calculators is abstract for people not familiar with the terms. Any input with the formula is greatly appreciated! Thx

I think you're idea of explaining this to people with the correct numerical answer is kind of hopeless. Whether you are a cop or a physicist, people just don't have a great intuitive visceral understanding of impact energy, impulse, momentum, etc.

So, I think you are better off with analogies. Suppose you put a beam out of the second story window of a building (22 ft. high) and hung a bag of cement (90 Lb) from it at person/dog level. Then you wrap the cement in foam to make it about as squishy as a dog. Then swing that bag up to horizontal (with the rope always straight) and ask if anyone would volunteer to catch it when you let it swing down. That is roughly comparable.

edit: However, I don't think that's how dogs actually hit the decoy, it's nearly always off axis with rotational motion to reduce the impact forces (less force x longer duration) and distribute the energy between translation and rotation. So this analogy is an exaggeration. Ask your decoy: would you rather have the dog bite your arm or slam directly into your chest?

 

[DougCoMo]

It's easy to come up with a number like that, but what does it mean? You could go back to the physics teacher and ask what if it's a 75 lb rock? Why wouldn't you get the same answer?

At what point in the calculation does the fact that it's a dog and not a rock become relevant?

For my purposes, never.

 

[DougCoMo]

It seems to me that a direct impact between the centers of mass of running-jumping dog and static suspect seldom occur.
By watching these slow motion videos, we can see that the dog goes for one of the arms of the subject, which makes it use some of its impulse to jump in order to bite higher.
Then, a natural rotation of the suspect's body happens, which dramatically reduces the efect of the impact on the balance of the subject.

The fun part starts when the impulse of the dog takes it to the opposite side of the original trajectory and pulls the entire body of the suspect via its canines bitting his arm.
I believe that it is at that point when the dogs can break canines and have disc injuries, because the material of the protecting gear does not tear and yield as flesh and skin do.
The training subject does not feel pain either, therefore he does not go along with the impulse of the body of the dog, like a real suspect would do naturally trying to minimize pain and flesh damage.

In nature, the canines are strong if they can penetrate skin and flesh of an animal, because as the wolf shakes his head and body to tear flesh and to increase pain and bleeding, the associate huge forces are uniformly distribuited closer to the root of the canines.
Bitting on the surface of the protective gear makes the canines of a police dog more vulnerable because the forces that could break them are applied to the tip rather than the section closer to the gum (greater moment or torque applied at the root).
Same for the spine discs, if the spine is no aligned when traction or compression or twisting forces happen.

Perhaps adding some elasticity or yielding capability to the usual areas of bites in the garmet could aliviate dog's injuries, as well as training on more yielding movements to "pain" from the helpers.

All the above makes any calculation complicated.
A way to experiment about the effect of pure impact could be launching a bag of sand against the body of a standing man.
That man could try catching that bag off center and try holding it while it rotates to the opposite side.
Of course, that would leave the pain effect out of the situation.

The question is when there is impact, no spin, no change in original trajectory. The question is about a "jam" where a dog is injured. It's the #1 injury to professional dogs, there is no give and impact hits a man and causes him to move only about 18 inches. 75 pound dog 28 mph then impact.

 

[DougCoMo]

I think you're idea of explaining this to people with the correct numerical answer is kind of hopeless. Whether you are a cop or a physicist, people just don't have a great intuitive visceral understanding of impact energy, impulse, momentum, etc.

So, I think you are better off with analogies. Suppose you put a beam out of the second story window of a building (22 ft. high) and hung a bag of cement (90 Lb) from it at person/dog level. Then you wrap the cement in foam to make it about as squishy as a dog. Then swing that bag up to horizontal (with the rope always straight) and ask if anyone would volunteer to catch it when you let it swing down. That is roughly comparable.

edit: However, I don't think that's how dogs actually hit the decoy, it's nearly always off axis with rotational motion to reduce the impact forces (less force x longer duration) and distribute the energy between translation and rotation. So this analogy is an exaggeration. Ask your decoy: would you rather have the dog bite your arm or slam directly into your chest?

No, the off axis rotation is when the helper "catches"the dog with footwork and uses momentum and skill to return the dog to the ground. The body bite suit isn't intended for arm bites so it's a mute point. Its purpose is to teach the dog to bite in the "core". Chest to upper bicep, upper tricep to lat and also legs. It's designed to do what a bite "sleeve" can't do. Why wear 30 pounds of equipment to train a dog to bite your arm?

 

[A.T.]

The question is when there is impact, no spin, no change in original trajectory. The question is about a "jam" where a dog is injured.

But what is the question about that?

 

[DougCoMo]

But what is the question about that?

Thanks for the reply. I don't think it's a question that can be answered.