Weapond Physics

[Teryan]

I was about to go to bed one night and I saw a spider about to climb onto my bed, so I grabbed by staff and tried to poke it.... I missed and put a hole thru my wall. I did not go that hard :rolleyes: , so I got to wondering about what kind of power it had. Here are the results, math/physics :love: ppl plz check my work.
It was alot of fun to do a three sectional staff :)

First off Conversion Factors (It's been a wile)
3.28 ft in 1 m
2.56 cm in 1 ft
2.2 lb in 1 kg

Equations:
Force = Mass X Velocity, edit Force = Mass * Acceleration
Kenetic Energy (KE) = 1/2(Mass X Veloviety^2)
Pressure = Force / Area
Circle Area = PiR^2
Circle Cirumpus = 2PiR

Childs Staff (AKA Spider Killer).
I am using a foward poke with this caculation. This is assuming I am really trying to get some thing bigger than a spider. Using full force.
Dimensions: 5 ft. by 1/2 in. and 1/2 in. around (end)
Weight: 33.9 oz, or .963 kg
Striking area of the end of the staff (area that makes contact with object): Pi(1/4 in)^2 = .196 in. and to get it into metric, .196/ 2.56 = .072 cm.
Assuming I use 40% of my body weight: 123 lbs * .4 = 49.2 lbs x (1/2.2kg) = 22.36 kg
Pokeing speed = 8 ft/sec * (1ft/3.28m) = 2.44 m/sec
Force = 23.32 kg * 2.44 m/s = 56.88 kg/m/s
KE = 1/2(23.32 kg * 2.44^2) = 69.4189 J
Pressure = 56.88 m/s / .077 cm^2 = 738.701 m/cm^2 * (3.28 m/ 1ft) = 2422.94 in/cm^2 * (1 in / 2.54 cm) = 953.913 lbs/ in.^2 aka PSI

I have also got some stats on my traning three sectional (has foam all around it) and a real one. I need to check some things on it, and if you guys like I will post it to. I would also like to check the Forec, KE, Pressure of other weaponds, so if you guys have any weaponds that you want checked, just give me some of the stats I mentioned above.

[Sphyerion]

might want to take into account technique etc. there are variations of every weapons so weight is hard to say...

you should notice that depending on how you use a weapon.. the leverage from your joints needs to be taken account of.

very interesting stuff though...

[shinguards]

that's pretty extreme, isn't it..? taking a staff to kill a spider...?

[Teryan]

that's pretty extreme, isn't it..? taking a staff to kill a spider...?

He caught me off guard, and I was tired, just wanted to get to bed. :) Hey I did't use my 6ft heavy staff for him.

you should notice that depending on how you use a weapon.. the leverage from your joints needs to be taken account of.

I know things are different from weapond to weapond, but the main consern should be the weapond it's self. The boy can add (or subtract) to the weaponds veloicty and can put more mass behind it. Notice I used only one technique with that (a poke) and their are many other that can be caculated (fiture eights, spinning, one side grab and swing). I'm not shure what you mean by joins. Do you mean that if you wrap some one wrist up in a nunchuka(sp) it and pull? Or that (body) joints add veloicty?

Any one know the PIS it takes to break bones? I have heard of these:
38 psi for ribs,
1400 psi for skull (from the Discovery channel's XMA thing)
8 for psi the big toe.

[Capt Ann]

Don't mean to be pedantic, but....

Force = mass x acceleration (not velocity)
When converting in^2 to cm^2, you must divide by 2.54 cm/in twice

Perhaps next time you can dangle the staff from a cord of known length, then release the staff from 0 velocity starting point, use the change in height of the staff from beginning to end, and the gravitational constant to calculate the change in potential energy, then use the Law of Conservation of Energy to derive the final velocity from the calculated kinetic energy, then back out the acceleration to get the force and pressure............

Or you could just buy a can of Raid, like the rest of us ;)

[dustIn credible]

if you want to do a live test not just numbers use a watermellon for the human skull.......it has almost exactly the same tensile strength as the skull and if you can find on that weights about the same use that....makes you stop and think about how easily demolished the skull really is and how much force it can really take.......watermellons dont have however the elasticity most humans skulls do so therefore cant be completely accurate but 99% most of the time........your skull might not feel like it but it actually does have alot of "give" to it making it able to withstand forceful blows but watermellons you can create a pretty accurate representation if you all want to expierement with different techniques, weapons, ect...

[LilBunnyRabbit]

Just a quick point, if you're poking with the staff then the only thing that makes any difference is the size of the impact area of the staff, and the length is irrelevant. If you're swinging the staff then moments will apply, but not with a poke.

[Kwajman]

Your poking speed is the greatest variable I think. But thats not constant. Anyway, I am impressed and also concerned that you have this much free time to figure out. But I do think you get an A for your work... :D

[Sphyerion]

well, when I say joints.. joints are equivalent to levers... levers have different calculations i believe. just a thought.. i could very well be wrong..

care to do the calcs for the average Katana?

[Teryan]

care to do the calcs for the average Katana?

Shure. Just give me the acceleration of the Katana, the mass that you are striking with, and the area of the striking area of the katana. I'm shure the area of the Katana is going to be extremely small, so make a educated guess guess to what it is.

if you want to do a live test not just numbers use a watermellon for the human skull.......it has almost exactly the same tensile strength as the skull and if you can find on that weights about the same use that

Really? I had no idea. I was told a pineapple was equal to the human skull. I'd like to give it a try some time, and if i do i'll post some vids of it.

and the length is irrelevant.

Yep, the lenght is irrelevant for a poke, but I was also doing some caculations on swinging too.

well, when I say joints.. joints are equivalent to levers... levers have different calculations i believe. just a thought.. i could very well be wrong..

No, beacuse the added leverage (joints) added to the speed of the weapond. It could have it's own caculations if you really wanted, but it would not tell us any thing new that we did not know. Maby it would, it would tell us how mutch veloicty we are puting in from our arms, vs. our legs (assuming you step foward on your strike).

Thanks for the responses guys.

[Teryan]

Don't mean to be pedantic, but....

Force = mass x acceleration (not velocity)
When converting in^2 to cm^2, you must divide by 2.54 cm/in twice

Perhaps next time you can dangle the staff from a cord of known length, then release the staff from 0 velocity starting point, use the change in height of the staff from beginning to end, and the gravitational constant to calculate the change in potential energy, then use the Law of Conservation of Energy to derive the final velocity from the calculated kinetic energy, then back out the acceleration to get the force and pressure............

Or you could just buy a can of Raid, like the rest of us ;)

Your right about the Force = mass * accleration. I ment acceleration, not veloicty.

The conversition factor :confused:, why do i half to do it twice? would't that be in to cm (1st time,) then cm to cm (2nd time)?

As for droping the staff, the acceleration is going horizontal not vertical. It also does not take into account how fast I poke, only how fast gravity falls (9.8 m/s^2). All of that would be great if we wanted to caculate the Force, KE, and PSI of the staff if it was just falling, no human behind it.
Thanks for the thoughts.

And I am looking into the can of raid. :)

[Mort]

Where did the spider go if you missed him?

[Teryan]

Where did the spider go if you missed him?

In my bed some where. Just proves their is a weapond for every job, this time it was raid. Good thing I did not have a katana :D
Those of you that are not into the math thing, just look at the colored numbers.

Same equations
Same conversions

Three Sectional Staff:
Dimensions (per section. total: three sticks, four bearings, two cahins): 25 in (82cm) stick, 1(3/4) in. (2.46 cm) bearing, 2.5 inch (8.2 cm) chain.
Also using arm for added lenght: 314.88 cm.
Roations: Two per second.
Striking area = 2 sq cm or .002 sq m(it would be extremely lower if you hit with an edge)

Mass: 3.2 oz per section. Tottal is 6.4 oz. for staff. and with body it is 23.98 kg.
Assuming using 20% body weight: 123 lbs. / 2.2 kg. = 55.9 kg
First 3/4 of circle rotation is "Crusing". Getting ready to strike. Two of these circle a second.
Last 1/4 of circle rotation is where acceleration is added along with the extenion of arms and the rest of the staff. The last fourth of the circle is accelerated 5-8 times.

First 3/4 of circle roation (Crusing):
Using the first section and the chain and one ball baring:
2Pi(56.5 in.) = 355 in. (one full circle) * (3/4) = 266.3 in. (3/4 of a circle) or 873.43 cm.
Last 1/4 of cirlce roation (the strike):
This is using my arm and two and a half section of the staff.
2Pi(96 in.) = 903.19 in. (one full circle) * (1/4) = 150.79 in. or 383.0066 cm.
All together the distance travled is:
266.3 + 150.79 = 417.09 in. or 1059 cm.

Two rotations a second:
This is to caculate the crusing speed
2Pi(56.5) = 355 in/2 seconds * 2 = 710 in/sec * (1ft/12 in.) = 59.17 ft/sec (rotation a second) * (3/4) = 44.379 ft/sec or 13.52 m/s
This is to caculate the striking speed
13.52 m/s * 5 = 57.644 m/s or 221 ft/sec
Acceleration = 67.644 - 13.52 = 54.12 m/s
Rember these caculations are on a good day (no injuries, stretched out, ideal conditinos) and they are not the most exact thigns in the world.

Force = 23.98 kg * 54.12 m/s = 1297.89 kg/m/s^2
KE = 1/2(23.98 * 54.12m/s^2) = 35118.403 J
Pressure = (1297.89 kg/m/sec^2) / (.002 sq m) = 648,945 kg/sq m/sec * (1m/3.28ft) = 197,849.08 kg/sq ft/sec * (1ft/12in) = 16,487.423 Kg/ sq in /sec * (2.2lbs / 1kg) = 36,272.33 lbs/ sq in/ sec :eek: :eek: :eek:
Now lets rember that flexable weaponds bouce back (and it hurts like hell when you dont expect it), so they spend less than a second on their. The harder you go the less time their will be on their, but you will be hitting it harder.
Lets say that the staff remains on the target for an 1/8 of a second (I have no idea how long it really does, just guessing here). So:
36,272.33 lb/ sq in/ sec * (1/8 sec) = 4,534.04 lb/ sq in/ (1/8) sec
That is enought to break someone skull at lest three time over :Angel:

[Reiki]

Or you could just buy a can of Raid, like the rest of us ;)

hey! it's more fun squashing them than nuking them! :D

I make it my job to squash every white tailed spider I find in the house and outside. Their bite is poisonous and they are nasty little SOB's that go for you if u move anywhere near them... Unfortunately they eat other spiders so this makes them more toxic.

A girl at my son's school got bitten by one 18mths ago and nearly had to have her leg amputated because of it.

[Sphyerion]

geez... that's one hell of a spider. =/

[Capt Ann]

The conversition factor :confused:, why do i half to do it twice? would't that be in to cm (1st time,) then cm to cm (2nd time)?

for inches squared (in^2), you have to divide by (2.54 cm per inch) squared to get cm squared.

As for dropping the staff, the acceleration ..........I was just suggesting you tie the staff to a string and turn it into a big pendulum.....just because it would be easier to calculate how fast it's going at "Tsi" (time of spider impact)

I still think the Raid idea's your best bet.

[xplasma]

I was about to go to bed one night and I saw a spider about to climb onto my bed, so I grabbed by staff and tried to poke it.... I missed and put a hole thru my wall. I did not go that hard :rolleyes: , so I got to wondering about what kind of power it had. Here are the results, math/physics :love: ppl plz check my work.
It was alot of fun to do a three sectional staff :)

First off Conversion Factors (It's been a wile)
3.28 ft in 1 m
2.56 cm in 1 ft
2.2 lb in 1 kg

Equations:
Force = Mass X Velocity, edit Force = Mass * Acceleration
Kenetic Energy (KE) = 1/2(Mass X Veloviety^2)
Pressure = Force / Area
Circle Area = PiR^2
Circle Cirumpus = 2PiR

Childs Staff (AKA Spider Killer).
I am using a foward poke with this caculation. This is assuming I am really trying to get some thing bigger than a spider. Using full force.
Dimensions: 5 ft. by 1/2 in. and 1/2 in. around (end)
Weight: 33.9 oz, or .963 kg
Striking area of the end of the staff (area that makes contact with object): Pi(1/4 in)^2 = .196 in. and to get it into metric, .196/ 2.56 = .072 cm.
Assuming I use 40% of my body weight: 123 lbs * .4 = 49.2 lbs x (1/2.2kg) = 22.36 kg
Pokeing speed = 8 ft/sec * (1ft/3.28m) = 2.44 m/sec
Force = 23.32 kg * 2.44 m/s = 56.88 kg/m/s
KE = 1/2(23.32 kg * 2.44^2) = 69.4189 J
Pressure = 56.88 m/s / .077 cm^2 = 738.701 m/cm^2 * (3.28 m/ 1ft) = 2422.94 in/cm^2 * (1 in / 2.54 cm) = 953.913 lbs/ in.^2 aka PSI

I have also got some stats on my traning three sectional (has foam all around it) and a real one. I need to check some things on it, and if you guys like I will post it to. I would also like to check the Forec, KE, Pressure of other weaponds, so if you guys have any weaponds that you want checked, just give me some of the stats I mentioned above.


Air Resistance, you don't live in a vaccuum.

[Teryan]

for inches squared (in^2), you have to divide by (2.54 cm per inch) squared to get cm squared.

But would't that only be if I squared the in.? I dont, the in^2 is only saying that it covers an area.

Air Resistance, you don't live in a vaccuum.


Ok, any idea how to caculate that? I dont think the effect would be so great on the staff as it would be on the three sectional. I have also heard that air resistance does not make a huge difference.

[Capt Ann]

Quoting Capt Ann:
for inches squared (in^2), you have to divide by (2.54 cm per inch) squared to get cm squared.

But would't that only be if I squared the in.? I dont, the in^2 is only saying that it covers an area.

Trust me, it works the same way.

Imagine you have nine square inches in area.......three inches down by three inches across. That would be 3 x 2.54 cm/inches down by 3 x2.54 cm/inches across, to get the same area in square centimeters. Same as 9 square inches x (2.54 cm/in) squared.

[Teryan]

I checked it mathmatically and your right. I dont have alot of time any more (spring break over :( ) so, it may be a wile before i get to fix it.

Thanks for your help.

[Fudotai]

Force is one thing, but as you say, you missed. if you would have used less force and more control, perhaps there would have been one more dead arachnid...
"Never use more force than is necessary"
The Bubishi