Q: Power of dynamite?
John Self
I'm writing an article about the kinetic energy stored in traffic on
motorways, and I'd like to make the comparison with the amount of energy
stored in "a stick of dynamite", since I think that it will have a
certain shock value. Along the lines of "...a 38 tonne truck travelling
at 60 mph has stored kinetic energy equivalent to a 20 oz stick of
dynamite..."
Is there anyone who can give me some help here? I'm sure that there will
be many definitions of "stick of dynamite", and I'm not quite sure how
the energy of dynamite can be equated to E=MV terms, but any information
would be much appreciated.
Thanks
John
(My company sponsors the bandwidth - the opinions are mine)
Matti Pusa
:Hi,
:
:I'm writing an article about the kinetic energy stored in traffic on
:motorways, and I'd like to make the comparison with the amount of energy
:stored in "a stick of dynamite", since I think that it will have a
:certain shock value. Along the lines of "...a 38 tonne truck travelling
:at 60 mph has stored kinetic energy equivalent to a 20 oz stick of
:dynamite..."
Let's see...
60 mph = 97 kph = 27 m/s, 20 oz = 0,570 kg = 570 g, 38 tons is 38 000 kg
(unless you mean 38 thousand pounds). E = 1/2 m v^2.
Truck's kinetic energy is: Etruck = 1/2 * 38 000 kg * (27 m/s)^2 = 13 851
000 J = 13,9 MJ.
If I recall correctly nitroglyserin holds 1500 calories of energy per gram.
1500 cal = 6290 J. Dynamite might be 50 / 50 mixture of nitroglyserin and
diatomious earth, so the energy might be for example 3145 J per gram.
Stored energy of stick of dynamite is: Enitro = 3145 J / g * 570 g = 1 794
000 J = 1,79 MJ.
13,9 MJ / 1,79 MJ = 7,7
Conclusion: About eight 20 oz sticks of dynamite would be needed to produce
the same energy that 38 ton truck, moving 60mph, has. So if we could convert
that truck's hole kinetic energy into heat and other suitable forms of
energy within say one thousand of a second, the effect might be a lot the
same as that of 8 sticks of dynamite.
Some interesting comparements:
If a man, weighing 100 kg, was dropped from an airplain so that he's speed
reaches the maximum possible, about 250 kph, and he would crash into ground,
there would be about one tenth of energy of a dynamite stick converted from
kinetic energy into other energy formes in a very short time.
Or if a meteor, weighing 1kg (about the size of a tennis ball) and
travelling 40 kilometers per second, hit the ground without losing any of
it's speed in the atmosphere: The energy would be 800 MJ, the same as for
about 1,5 tons of TNT. But, if the meteor was actually an asteroid, called
Ceres (the biggest known in our solar system), which weighs more than 10^21
kg (diameter almost 1000 km) and has a travel speed of about 9 km per
second: E = 4 * 10^22 MJ = 10^13 megaton explosion (10 000 billion one
megatons nuclear weapons at the same time, fireworks display in the real
meaning of the word). The explosion which probably killed the dinosaurs was
millions of times less powerfull. Surprisingly it only takes less than 10
second for our sun to produce the corresponding energy by fusion.
:Is there anyone who can give me some help here? I'm sure that there will
:be many definitions of "stick of dynamite", and I'm not quite sure how
:the energy of dynamite can be equated to E=MV terms, but any information
:would be much appreciated.
I don't know if this helped, but I had fun calculating the energys.
:Thanks
:
:John
Matti Pusa mapus<at>sci<dot>fi
Micha und Olli
>
It´s E=1/2 mv² !Micha !
John Self
>
> John Self wrote in message <35BC5C...@nospam.uk>...
> :Hi,
> :
> :I'm writing an article about the kinetic energy stored in traffic on
> :motorways,
> Let's see...
>
> 60 mph = 97 kph = 27 m/s, 20 oz = 0,570 kg = 570 g, 38 tons is 38 000 kg
> (unless you mean 38 thousand pounds). E = 1/2 m v^2.
>
> Truck's kinetic energy is: Etruck = 1/2 * 38 000 kg * (27 m/s)^2 = 13 851
> 000 J = 13,9 MJ.
>
> If I recall correctly nitroglyserin holds 1500 calories of energy per gram.
> 1500 cal = 6290 J. Dynamite might be 50 / 50 mixture of nitroglyserin and
> diatomious earth, so the energy might be for example 3145 J per gram.
> Stored energy of stick of dynamite is: Enitro = 3145 J / g * 570 g = 1 794
> 000 J = 1,79 MJ.
>
> 13,9 MJ / 1,79 MJ = 7,7
>
> Conclusion: About eight 20 oz sticks of dynamite would be needed to produce
> the same energy that 38 ton truck, moving 60mph, has. So if we could convert
> that truck's hole kinetic energy into heat and other suitable forms of
> energy within say one thousand of a second, the effect might be a lot the
> same as that of 8 sticks of dynamite.
more snip-type surgery
> I don't know if this helped, but I had fun calculating the energys.
>
> :Thanks
> :
> :John
>
> Matti Pusa mapus<at>sci<dot>fi
Thanks for that information Matti, it helped me enormously. I'm glad you
also had fun doing the calculations.
Best regards
John
Roger Riordan
>I'm writing an article about the kinetic energy stored in traffic on
>motorways, and I'd like to make the comparison with the amount of energy
>stored in "a stick of dynamite", since I think that it will have a
>certain shock value. Along the lines of "...a 38 tonne truck travelling
>at 60 mph has stored kinetic energy equivalent to a 20 oz stick of
>dynamite..."
>
>Is there anyone who can give me some help here? I'm sure that there will
>be many definitions of "stick of dynamite", and I'm not quite sure how
>the energy of dynamite can be equated to E=MV terms, but any information
>would be much appreciated.
The answer I received to a similar question some time ago was:
One megaton = 4.17 x 10^15 joules, or the energy released
by 47 grams of matter being converted completely into energy.
So 1 kg of TNT will give 4.17 MJoules. Guess dynamite is roughly
similar.
60 mph ~ 27m/s, & KE = 0.5MV^2 =
0.5 * 38E3 * 730 = 13.8MJ ~ 3kg of TNT
Roger Riordan
Roger Riordan
>John Self wrote in message <35BC5C...@nospam.uk>...
.......................................
>Some interesting comparements:
................
>Or if a meteor, weighing 1kg (about the size of a tennis ball) and
>travelling 40 kilometers per second, hit the ground without losing any of
>it's speed in the atmosphere: The energy would be 800 MJ, the same as for
>about 1,5 tons of TNT.
Most of your figures agree reasonably well with mine, but this
doesn't. According to the figures I was given 1T of TNT -> 4.2GJ, so
800MJ ~ 0.2T of TNT
>But, if the meteor was actually an asteroid, called
>Ceres (the biggest known in our solar system), which weighs more than 10^21
>kg (diameter almost 1000 km) and has a travel speed of about 9 km per
>second: E = 4 * 10^22 MJ = 10^13 megaton explosion
Much worse discrepancy here; I get 10^7MT. Puny by comparison :-(
(10 000 billion one
>megatons nuclear weapons at the same time, fireworks display in the real
>meaning of the word). The explosion which probably killed the dinosaurs was
>millions of times less powerfull. Surprisingly it only takes less than 10
>second for our sun to produce the corresponding energy by fusion.
>
>:Is there anyone who can give me some help here? I'm sure that there will
>:be many definitions of "stick of dynamite", and I'm not quite sure how
>:the energy of dynamite can be equated to E=MV terms, but any information
>:would be much appreciated.
>
>
>I don't know if this helped, but I had fun calculating the energys.
>
>:Thanks
>:
>:John
>
>Matti Pusa mapus<at>sci<dot>fi
>
Roger Riordan
Matti Pusa
:"Matti Pusa" <nospa...@spam.no> wrote:
:
:>John Self wrote in message <35BC5C...@nospam.uk>...
:.......................................
:>Some interesting comparements:
:................
:>Or if a meteor, weighing 1kg (about the size of a tennis ball) and
:>travelling 40 kilometers per second, hit the ground without losing any of
:>it's speed in the atmosphere: The energy would be 800 MJ, the same as for
:>about 1,5 tons of TNT.
:
:Most of your figures agree reasonably well with mine, but this
:doesn't. According to the figures I was given 1T of TNT -> 4.2GJ, so
:800MJ ~ 0.2T of TNT
According to my source, TNT holds 1295 cal/g, which is (1 cal = 4,187
joules) 5422 J. So 1000g of TNT is 5,422 MJ, 1000kg of TNT is 5,422 GJ
(almost the same as your figure, Roger) . 800 MJ compares to: 800 / 5422 =
0,15; 150kg of TNT - Your right, I was careless with my calculations :-(
:>But, if the meteor was actually an asteroid, called
:>Ceres (the biggest known in our solar system), which weighs more than
10^21
:>kg (diameter almost 1000 km) and has a travel speed of about 9 km per
:>second: E = 4 * 10^22 MJ = 10^13 megaton explosion
:
:Much worse discrepancy here; I get 10^7MT. Puny by comparison :-(
I should have writen everything up here in the first place, it wouldn't have
been any more work and it would have pretty much eliminated the errors that
I have made, my fault. The WIN95-calculater I used hasn't got any finenesses
;-)
E = 0,5 * 10^21 kg * (9 000 m/s)^2 = 4,05 * 10^28 J = 4,05 * 10^19 GJ. As I
earlier calculated, 1 T TNT has about 5 GJ of energy, so one kiloton is 5 *
10^3 GJ, one megaton 5 * 10^6 GJ. 4 * 10^19 GJ / 5 * 10^6 GJ = 8 * 10^12,
about 10^13 MT.
I didn't find anything wrong, but that doesn't mean there may not be
anything wrong. Please tell where I went wrong.
The funny thing in this energy-calculation thing is, that the speed is much
more important factor than the mass. Traffic accidents are an unfortunate
live example of this (for example (I'm not sure if this is complitely
correct) when your speed increases from 40 kph to 50 kph, the braking
distance increases 100 percent :0). Example: 1kg object travelling at light
speed, 300 000 km/s, has 4,5 * 10^7 GJ of energy, thus having the power of
about 10 megatons. (Sorry I don't have time to write these calc. here.) You
understand the huge powers of nuclear fusion and fission when doing these
calculations.
<snip>
Roger Riordan
...........................................................
>:>But, if the meteor was actually an asteroid, called
>:>Ceres (the biggest known in our solar system), which weighs more than
>10^21
>:>kg (diameter almost 1000 km) and has a travel speed of about 9 km per
>:>second: E = 4 * 10^22 MJ = 10^13 megaton explosion
Sorry; you were right. I didn't notice the M before the J - not very
good practise to mix two different types of multiplier; 10^28J would
be better. However I note that I had followed your lead in talking
about 10^XXmegaton, so I was guilty too :-)
>The funny thing in this energy-calculation thing is, that the speed is much
>more important factor than the mass. Traffic accidents are an unfortunate
>live example of this (for example (I'm not sure if this is complitely
>correct) when your speed increases from 40 kph to 50 kph, the braking
>distance increases 100 percent :0).
Assuming your brakes can provide constant deceleration A, stopping
distance S = V^2/A, so going from 40 to 50 increases distance by
1.25^2 = 1.56 or 56%. Real situation is probaly a bit worse than
this, as the assumption about brakes is probably a bit optimistic, but
I suspect the 100% figure was devised by the safety authorities to
scare you.
> Example: 1kg object travelling at light
>speed, 300 000 km/s, has 4,5 * 10^7 GJ of energy, thus having the power of
>about 10 megatons. (Sorry I don't have time to write these calc. here.) You
>understand the huge powers of nuclear fusion and fission when doing these
>calculations.
Relativity states that it takes infinite energy to accelerate a body
having finite rest mass to the speed of light. However if you can
devise a way of converting mass into energy (such as reacting it with
an equal mass of antimatter :-) Einsteins E = MC^2 applies, and 1 kg
will give about 9 * 10^16J, or about the same as 20MT. And this is
enough to supply 100MW for 30 years.
Roger Riordan