How do I calculate the G Force in this certain situation?

[Ishay]

Hello,

I have a question related to a specific situation.
I've been playing a game called Overwatch and while I'm not the smartest there is in physics I'd like to understand it better.

In Overwatch there is a character named Roadhog he wields a shotgun and a hook.
Using this hook he can launch it towards a certain enemy at the maximum range of 20m at the speed of 64.5m/s (he reaches the maximum of 20m in 0.31 seconds),Roadhog grabs his target and yanks him/her towards himself at the speed of 40m/s over 0.5 seconds no matter the size(he can even pull the enemy Roadhog that weighs 550lbs or 250kg).
After I gave you all of the information I know I'd like to know 'What is the g force a person would experience if the hook hits him at 232.2 km/h(64.5m/s) while neglecting friction'

*Note that as soon as Roadhog throws the hook there is no acceleration(the minute he throws it the velocity stays the same)

*Note that I am talking about the hook itself and not if the chain itself can survive such pressure when talking about tensile strength.

Hooks dimensions:
Height: 50cm
Depth:4cm
Width:19cm
Mass: Around 4kg

This is not..homework of some sort its just curiosity.
F = M * A
Ek = _1/2 M*V²
A = v_f - v_i / time

Thank you.

 

[BvU]

Hello Ishay,

Well, no physics went into the design of this feature. You ever seen a chain that doesn't sag ? Perhaps the thing is weightless... .

But you were asking about the hook. Its kinetic energy is enormous: 4 kg at 64 m/s is around ($\ {1\over 2} mv^2\$) 8000 Joule. So to stop it in, say, 0.5 m (the clip shows the target flexing backwards), the victim exerts 16000 N (8000 J/ 0.5 m) . If he weighs 100 kg that's 16 g forces (someone jumping on your shoulder from four floors up, feet first). Poor guy's arm (plus shoulder, probably) would flail off.

 

[Ishay]

Hello Ishay,

Well, no physics went into the design of this feature. You ever seen a chain that doesn't sag ? Perhaps the thing is weightless... .

But you were asking about the hook. Its kinetic energy is enormous: 4 kg at 64 m/s is around ($\ {1\over 2} mv^2\$) 8000 Joule. So to stop it in, say, 0.5 m (the clip shows the target flexing backwards), the victim exerts 16000 N (8000 J/ 0.5 m) . If he weighs 100 kg that's 16 g forces (someone jumping on your shoulder from four floors up, feet first). Poor guy's arm (plus shoulder, probably) would flail off.

Thank you for the explanation. (I took physics class only in 9th grade for one year so if I make mistakes sorry.(currently 11th))
When we talk about the hook hitting the person. the person would stop the hook in the distance of 0.1m or even less.
So If I understood well,
E = _1/2(4kg * 64.5m/s²) = 8320.5 J
The chain was traveling at 64.5m/s with the energy equivalent 8320.5 J
The average weight of an adult human being is 62 kg(Wikipedia)
The object, in this case the victim would resist 83205N (8320.5 J / 0.1m),the amount of energy the person experiences is a little over 25 times the force required to have a 25% chance of cracking an average human rib(according to this article Paragraph 2).
With this amount of energy its safe to say that some of your ribs if not all would break and puncture his lungs, correct me if I'm wrong in my calculations.
And one last question when you calculated the g force

If he weighs 100 kg that's 16 g forces (someone jumping on your shoulder from four floors up, feet first). Poor guy's arm (plus shoulder, probably) would flail off.

how did you calculate it?

 

[BvU]

how did you calculate

The kinetic energy of approximately 8000 J can be dissipated over a distance of 0.5 m ( I really see the receiver flex backward considerably more than 0.1 m) by a force of 8000 J / 0.5 m = 16000 N. I came to 16 g by estimating the guy at 100 kg, so a scale would read 100 kg x 9.8 m/s² $\approx$ 1000 N.

The formulas I use are pretty standard (first and last section).

There is a lot more to be said about these two seconds: there is also a Newton's third law: action = $-$ reaction. The shooter person Looks as if he (she?) is throwing the thing without any effort.
And if in fact the stopping action isn't performed by the receiver but by the sender at the other end of the chain (the weapon is a hook, after all), then that would produce a jerk hefty enough to throw him her over or displace him/her over a considerable distance.

Look at shot throwing. (Coincidentally a women's shot for throwing is 4 kg as well.) They have to spin elaborately to make the thing describe a curved arc (a parabola). Your shooter doesn't seem to recoil -- or even have to make an effort -- to impart the 8000 J of kinetic energy to the hook !

Two more things:
1.

the amount of energy the person experiences is a little over 25 times the force required

You can not, ever, say that an energy is a number of times a force. The dimension of energy is different from the dimension of a force (by one factor of length). But you didn't actually do that: you compared the 80000 N you calculated on the basis of 0.1 m stopping distance with the 3300 N in the ninth paragraph in your link.
(very nice: 25% chance at 3300 Newton means ... what chance at 80000 N ?)

2. PF doesn't appreciate asking subject questions in private conversations. That's what the threads are for. So others can benefit from the answers too. And if I make errors or false claims, others can put me right, etc..

 

[Ishay]

The kinetic energy of approximately 8000 J can be dissipated over a distance of 0.5 m ( I really see the receiver flex backward considerably more than 0.1 m) by a force of 8000 J / 0.5 m = 16000 N. I came to 16 g by estimating the guy at 100 kg, so a scale would read 100 kg x 9.8 m/s² $\approx$ 1000 N.

Thanks for answering,pretty of stupid of me to forget the first formula I've learnt.
On a side note,to reply to your statement of

0.5 m ( I really see the receiver flex backward considerably more than 0.1 m)

I've made this to show that its not 0.5m.
The mechanic of the hook is to go behind the character and strip him of his abillity to move while closing the gap between the two.
It goes behind him,I was theorising on the effects a human would experience if he would've get hit by it,not hooked.
Even if you're somehow not convinced..its just a visual effect in the game to 'flinch', its meant to let other players know the player got hooked.

Two more things:
1. You can not, ever, say that an energy is a number of times a force. The dimension of energy is different from the dimension of a force (by one factor of length). But you didn't actually do that: you compared the 80000 N you calculated on the basis of 0.1 m stopping distance with the 3300 N in the ninth paragraph in your link.
(very nice: 25% chance at 3300 Newton means ... what chance at 80000 N ?)

630.340909%,83205J

2. PF doesn't appreciate asking subject questions in private conversations. That's what the threads are for. So others can benefit from the answers too. And if I make errors or false claims, others can put me right, etc..

I didn't want to double post and another reason is I didn't want to look like an annoying guy by 'demanding' an answer.

 

[BvU]

I've made this to show that its not 0.5m.
The mechanic of the hook is to go behind the character

Clearly. So this character flexes towards the hook thrower.
Whereas the poor victim in post 1 flexes away from the thrower at first. So much for consistency .
As I said,

no physics went into the design of this

630.340909%,83205J

What does a probability of over 100% mean to you ?
(Compare: if you throw a die, the chance to get a six is 1/6. If you throw 2, 3, 4,... etc. What is the chance to throw a six if you throw 12 dice ?)
I also like the precision of your answer; you certain it isn't 630.340911% ?

I didn't want to double post and another reason is I didn't want to look like an annoying guy by 'demanding' an answer

No problemo.

 

[Ishay]

Clearly. So this character flexes towards the hook thrower.
Whereas the poor victim in post 1 flexes away from the thrower at first. So much for consistency .

You judge by different prespectives. I don't know about you but basic logic tells me a 6v6 PvP(I hope I don't have to tell you what PVP is)
needs to be fair and there should not be any reason for different characters to have different animations when hooked except hitboxes.
You're just taking a blind shot.

What does a probability of over 100% mean to you ?
(Compare: if you throw a die, the chance to get a six is 1/6. If you throw 2, 3, 4,... etc. What is the chance to throw a six if you throw 12 dice ?)
I also like the precision of your answer; you certain it isn't 630.340911% ?
No problemo.

Let me get this straight,you asked for the value when taking 80,000J into account and then when I give you the accurate value when its 83205J you bash my way of thinking. 13200 J is 100% in this case, why would you ask for 80,000J and then not...thats just plain stupid.
25% is 3300 then X is 83205,simple math.

I also like the precision of your answer; you certain it isn't 630.340911% ?

Pretty sure it isn't "630.340911" you should check your math, unless...and what do I know 8230.5 / 0.1 is 83205.0004