Date: 24/08/2020 22:12:43
From: wookiemeister
ID: 1609300
Subject: A tumbling object

let’s say you had an object travelling linearly through air , which then started to tumble. What percentage of the KE gets converted from linear KE to rotational KE ?

I think I’ve seen somewhere some shells are designed to deliberately tumble, so was thinking

Let’s say a shell left a barrel then started to tumble for whatever reason….

What’s the most likely rotational speed of the shell for a given starting velocity ? Rule of thumb ?

Eg shell travelling at 300m/s starts to tumble and reaches a maximum tumbling speed of x.

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Date: 25/08/2020 00:13:59
From: dv
ID: 1609337
Subject: re: A tumbling object

Surely it depends a lot on the shape…

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Date: 25/08/2020 07:29:18
From: wookiemeister
ID: 1609369
Subject: re: A tumbling object

dv said:


Surely it depends a lot on the shape…

Well yes. In this case a shell shape. Just something that popped into my head. I’ll head off into the internet to find something.

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Date: 25/08/2020 09:06:41
From: mollwollfumble
ID: 1609401
Subject: re: A tumbling object

wookiemeister said:


let’s say you had an object travelling linearly through air , which then started to tumble. What percentage of the KE gets converted from linear KE to rotational KE ?

I think I’ve seen somewhere some shells are designed to deliberately tumble, so was thinking

Let’s say a shell left a barrel then started to tumble for whatever reason….

What’s the most likely rotational speed of the shell for a given starting velocity ? Rule of thumb ?

Eg shell travelling at 300m/s starts to tumble and reaches a maximum tumbling speed of x.

As a fluid mechanics expert, I may just possibly be able to give you a ballpark estimate on that one.

But in the meantime, consider that air drag is a very big factor here. Nature wants to maximise air drag and, without spin stabilisation, a bullet will naturally want to travel broadside to the direction of travel. This is also a problem for rockets, which is why they need fins for stabilisation.

It’s not the energy transfer from kinetic energy to spin that slows the bullet, it’s the increased drag from the increased frontal area from flying sideways.

I would guess that the speed of spin is very close to the Strouhal Frequency. This frequency is 0.2 times the characteristic velocity (forward velocity of the bullet) times the characteristic length (length or diameter of bullet). From the frequency, mass and length, it is possible to calculate the rotational kinetic energy.

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Date: 25/08/2020 09:40:09
From: captain_spalding
ID: 1609421
Subject: re: A tumbling object

wookiemeister said:

I think I’ve seen somewhere some shells are designed to deliberately tumble, so was thinking

Projectiles are not designed to ‘tumble’ in flight.

Once they start doing that, any hope of accurate shooting goes out the window entirely.

What you’re probably referring to is the M193 5.56mm X 45 mm NATO cartridge. This was the original cartridge for the AR-15/M-16 rifle. That rifle originally had a 1-turn-in-14-inches rifling, which gave the M193 a relatively slow rate of spin.

That rendered the M193 only just stable in flight, particularly in cold air temperatures. When the bullet hit the target, it lost all stability, and it tipped over and slammed bodily into the target. As you can expect, this is bound to produce a much more severe wounding effect.

But, it was never supposed to tumble in flight

The slow spin rate didn’t do anything to help with accuracy, as the bullets would tend to yaw a bit after leaving the muzzle and the rifling was revised to 1-in12, and later to 1-in-7.

The later M855/SS109 cartridge had different propellant, and was considerably more stable in flight. However, both the M193 and M855 are meant to fragment on impact (otherwise, you’re just punching .22 size holes in things).

The M193 was not necessarily a bad cartridge, and it functioned quite well in early Ar-15s, M-16s, and M-16A1s. Indeed, it was far less likely than the M855 to cause cartridge feed problems, but that’s to do with gas pressures in the chamber and at the bolt face.

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Date: 25/08/2020 13:23:01
From: transition
ID: 1609554
Subject: re: A tumbling object

some more of my reading

https://en.wikipedia.org/wiki/Rifling

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Date: 25/08/2020 18:09:47
From: wookiemeister
ID: 1609721
Subject: re: A tumbling object

mollwollfumble said:


wookiemeister said:

let’s say you had an object travelling linearly through air , which then started to tumble. What percentage of the KE gets converted from linear KE to rotational KE ?

I think I’ve seen somewhere some shells are designed to deliberately tumble, so was thinking

Let’s say a shell left a barrel then started to tumble for whatever reason….

What’s the most likely rotational speed of the shell for a given starting velocity ? Rule of thumb ?

Eg shell travelling at 300m/s starts to tumble and reaches a maximum tumbling speed of x.

As a fluid mechanics expert, I may just possibly be able to give you a ballpark estimate on that one.

But in the meantime, consider that air drag is a very big factor here. Nature wants to maximise air drag and, without spin stabilisation, a bullet will naturally want to travel broadside to the direction of travel. This is also a problem for rockets, which is why they need fins for stabilisation.

It’s not the energy transfer from kinetic energy to spin that slows the bullet, it’s the increased drag from the increased frontal area from flying sideways.

I would guess that the speed of spin is very close to the Strouhal Frequency. This frequency is 0.2 times the characteristic velocity (forward velocity of the bullet) times the characteristic length (length or diameter of bullet). From the frequency, mass and length, it is possible to calculate the rotational kinetic energy.


Very useful

So an object now tumbling will have increased drag as you said

So now thinking about that whats a ballpark figure of drag for a tumbling let’s say artillery shell like shape?

Would there be a dramatic increase that slows the shell?

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Date: 25/08/2020 18:14:07
From: wookiemeister
ID: 1609723
Subject: re: A tumbling object

wookiemeister said:


mollwollfumble said:

wookiemeister said:

let’s say you had an object travelling linearly through air , which then started to tumble. What percentage of the KE gets converted from linear KE to rotational KE ?

I think I’ve seen somewhere some shells are designed to deliberately tumble, so was thinking

Let’s say a shell left a barrel then started to tumble for whatever reason….

What’s the most likely rotational speed of the shell for a given starting velocity ? Rule of thumb ?

Eg shell travelling at 300m/s starts to tumble and reaches a maximum tumbling speed of x.

As a fluid mechanics expert, I may just possibly be able to give you a ballpark estimate on that one.

But in the meantime, consider that air drag is a very big factor here. Nature wants to maximise air drag and, without spin stabilisation, a bullet will naturally want to travel broadside to the direction of travel. This is also a problem for rockets, which is why they need fins for stabilisation.

It’s not the energy transfer from kinetic energy to spin that slows the bullet, it’s the increased drag from the increased frontal area from flying sideways.

I would guess that the speed of spin is very close to the Strouhal Frequency. This frequency is 0.2 times the characteristic velocity (forward velocity of the bullet) times the characteristic length (length or diameter of bullet). From the frequency, mass and length, it is possible to calculate the rotational kinetic energy.


Very useful

So an object now tumbling will have increased drag as you said

So now thinking about that whats a ballpark figure of drag for a tumbling let’s say artillery shell like shape?

Would there be a dramatic increase that slows the shell?


The artillery shell shape is not not spinning ie no spin stabilisation.

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Date: 25/08/2020 18:16:39
From: wookiemeister
ID: 1609726
Subject: re: A tumbling object

wookiemeister said:


wookiemeister said:

mollwollfumble said:

As a fluid mechanics expert, I may just possibly be able to give you a ballpark estimate on that one.

But in the meantime, consider that air drag is a very big factor here. Nature wants to maximise air drag and, without spin stabilisation, a bullet will naturally want to travel broadside to the direction of travel. This is also a problem for rockets, which is why they need fins for stabilisation.

It’s not the energy transfer from kinetic energy to spin that slows the bullet, it’s the increased drag from the increased frontal area from flying sideways.

I would guess that the speed of spin is very close to the Strouhal Frequency. This frequency is 0.2 times the characteristic velocity (forward velocity of the bullet) times the characteristic length (length or diameter of bullet). From the frequency, mass and length, it is possible to calculate the rotational kinetic energy.


Very useful

So an object now tumbling will have increased drag as you said

So now thinking about that whats a ballpark figure of drag for a tumbling let’s say artillery shell like shape?

Would there be a dramatic increase that slows the shell?


The artillery shell shape is not not spinning ie no spin stabilisation.


As in when it was fired it was not spinning in the first place, it just started tumbling later on.

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Date: 27/08/2020 12:36:19
From: Ogmog
ID: 1610606
Subject: re: A tumbling object

Tumbling TRUMP Game

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