[Idol eyes]

Please can some one forward me the formula for working out K joules?
No reference paper work with me....
mental block,
Can not remember...
Is it...
m x f... oh fuck knows....

[dave]

if you're on about e=mc^2, it gives you energy in joules, so divide by 1000 to get kjoules.
m is mass in kilos
c is roughly 3x10^8 metres per second.

[Idol eyes]

Need to work out how many KJ certain weights strike on dymamic impacts from certain heights...
eg... 1.5kg falling 6mt's?
In fact if you can give me the formula, and the answer to this...
should be able to proceede!
Thanx!!!!
ps, is terminal velocity reached at 46 mts ish?
fucking wish I went to school!

[SA Chris]

Potential energy?

= m x g x h

[SA Chris]

So in your example U = 1.5 x 9.81 x 6m

(ignoring air friction effects)

terminal velocity of what though? Depends on the shape of the object and its frictional resistance as it travels through the air.

[Idol eyes]

1, teminal velocity of a 1.5 kg metal object (say a scoffold beam clamp/spanner)
2, A 90 kg person.
3, A three tonne I beam.
This is well out of my brain... but I im intested to gain info, since I have grazed the net and found lots of intresting phyisics stuff!
This is for a potential drooped object report as part of a ongoing rope access project! my laptop and paperwork/books were all lost during the snow saga last month at Aberdeen/Manchester airport!
Thanx SA.. I really am gonna start calling you the oracle!
   

[Idol eyes]

(dropped)

[lagerstarfish]

Bollox.
Stop confusing the lad, Chris. To work out the kJ you need to burn the object and see how much it heats up a known volume of water. It takes 4.18ish kJ to heat 1 litre of water by 1^oC. So in your case, Pat, you need to set fire to a 90 kg person under a bath of water and measure how much warmer it gets. You can make this more interesting by burning people you doon't like to heat up baths for attractive naked women.

 

[Idol eyes]

Even though your Methods is very interesting larger... and your banter would be great at a lecture SA... I might just wait till I get home and consult "Mark wrights "The Bible" rope access document!
Got the average figure... kind of all I needed really.
If I have learned something though its not to drop good looking women from a great height and guess dynamic impacts!
And not to bathe spanners and give them a "Chris Wentworth" bath to see how fat they might be!
Should be home today!
you could just give us the Layman principals like
MassxWeight divided by height or whatever it is!
really though thanx for that!

[Johnny Brown]

Chris' answer is spot on for your purposes Pat. With those three objects terminal velocity can be ignored, it wouldn't be reached off any structure in the UK bar possibly Emley moor (and on the person would depend on attitude, ie skydive stable or headdown pencil). I don't think Mark's book will give you any more info either, have found some good stuff on google previously though. Try looking on physics school sites.

Careful on those helis now, I bet there's a right fight to sit by the door now?

[SA Chris]

What JB said. I was just wearing my engineer hat when I started going on about air resistance. For hour porpoises (or dolphins for that matter) just use the formula. The spanner will bounce, the I beam will make a noise, the person will splatter. If you drop the I beam onto the person, however.....

[Johnny Brown]

What I'd be interested is being able to estimate is converting those kJ of potential energy into kN of impact force. I suppose there are too many variables but for instance falling 4m onto a FA lanyard we can calculate the Potential energy, and know the absorber will reduce the impact force to 6kN. What would it be without the absorber, ie a non-stretch lanyard? Is the potential energy the area under the curve in a force-time graph?

[lagerstarfish]

What would it be without the absorber, ie a non-stretch lanyard?

With a "non-stretch" lanyard the squishiness of the body in the harness and the tightening of any knots start to be significant contributers (along with the stretch of the "non-stretch" items) to the distance over which the energy will be absorbed and hence the resultant force. These are very difficult things to calculate on paper, so a physical test is the only way to get a reliable result.

The force generated by a falling object being stopped by a totally non-stretch system is infinity kN; which is quite a lot.

[Johnny Brown]

I appreciate all that, except that you can't get volunteers to do drop tests onto non-stretch lanyards. All the tests I've done have been on rigid masses. It'd be nice to have an idea roughly where between infinity and 6kN we are though.

Am I right in understanding the area under the force-time cuve as representing the energy though? I dropped physics after GCSE...

[erm]

No, the area under the line of force-time graph is the impulse.
The area under say a force-velocity graph would be energy.
If you compare the units of the two above, you will the first is kg*m*1/s (or Newton seconds) and the second is kg*m*1/s*s (which as it happens is energy).

Unforunately nice easy physics doesn't tend to translate well to solving real world problems.


Regards

[milksnake]

personally, id be more concerned about my spine dropping 4m onto a non strech system but what youd need to calculate to find out if it snaps is the average imact force.

[Johnny Brown]

Nice one erm. Reminds me of the old 'consider a spherical cow' joke.

[nic mullin]

The area under say a force-velocity graph would be energy.

The area under a force-velocity graph is power, area under a force-distance graph is energy word.

[nic mullin]

...and E = in case you were wondering.