# Impact and Detonation Calculations

• quatang
In summary, the impact of 1 ton of dynamite is approximately equivalent to dropping a 10,000 kg rock from 10 meters, which is about 0.28 kWh. However, 1 ton of dynamite/tnt is equivalent to approximately 4.184 gigajoules = 1162 kWh. Therefore, it is not possible to create an impact on the ground equivalent to 1 ton of dynamite by dropping a large boulder. A ~40 kiloton boulder would need to be dropped for a similar impact. The equation for kinetic energy is 0.5mv^2, and using this, a meteorite of around 5000-9000kg could hit the ground at a speed
quatang
Can someone help me to calculate the approximate impact on the Earth (as in a mining explosion), of 1 ton of dynamite.
And also the equivalent impact of say a 1 ton rock dropped from 10 meters.

Trying to estimate this for writing purposes, I very roughly got:

Impact of 1 ton of dynamite, Work = approx 1/2 megawatt.
Equivalent dropping a 1 ton rock 10 meters, approx 2/3 megawatt. Probably way off.

I didn't work very long at this and as I'll likely not be confident of getting the answer right, I thought I'd ask some people who would know how to do these calculations.

Appreciate any feedback.

quatang said:
Work = approx 1/2 megawatt.

quatang said:
dropping a 1 ton rock 10 meters, approx 2/3 megawatt
Watts and Megawatts measure power, not energy (work).

+1

Energy is measured in Joules.

The potential energy stored In a rock at height h is given by PE =mgh where m is the mass in kg, h the height in meters and g the acceleration due to gravity 9.8m/s/s.

You can look up the energy density of dynamite on Wikipedia.

@quatang ,

Look at my signature. Memorize that and you'll never make that mistake again.
Power is to Energy as Speed is to Distance

As @CWatters says, energy is pressured in joules. Electric power: 1 watt = 1 joule per second, 1 megawatt = 1 megajoule per second.

Thanks guys. To clarify my question, what I am after is, what mass of rock I would need to drop from say 20 meters, that would send an equivalent seismic wave into the ground to match a 1 ton dynamite blast.

Hope that's clearer. The physics units are pretty complex for a novice, so I'll avoid using them.

I only need a rough idea that is realistic.
Thanks for any help...

quatang said:
Thanks guys. To clarify my question, what I am after is, what mass of rock I would need to drop from say 20 meters, that would send an equivalent seismic wave into the ground to match a 1 ton dynamite blast.

Hope that's clearer. The physics units are pretty complex for a novice, so I'll avoid using them.

I only need a rough idea that is realistic.
Thanks for any help...

First of all, you cannot avoid using "physics units", because if you do, you'll be comparing apples to oranges.

Secondly, this link gives you the energy equivalent of 1 ton of TNT:

https://hypertextbook.com/facts/2002/AlexRoslyakov.shtml

Knowing this, will you be able to calculate the kinetic energy of impact of a rock of mass "m" from 20 meters up?

Zz.

Thanks guys. I think that's solved it.
I found that
A 10,000 kg rock, with 10 m drop, is about 1 MJ, which is about 0.28 kWh
And, 1 ton dynamite/tnt is approx 4.184 gigajoules = 1162 kWh

So there is no real possibility to create an impact on the ground by dropping a large boulder, equivalent to 1 ton of dynamite.

I estimate a ~40 kiloton boulder would need to be dropped. So this theme is out and back to the drawing board.

I'll check back tomorrow in case anyone can add anything to the above. Again thanks very for your comments.

Unfortunately gravity is surprisingly weak (although I find it's stronger early on Sunday mornings).

I ran some rough numbers for a meteorite impact. Googling suggests a meteorite of around 5000-9000kg might hit the ground at say 1km/s as it's slowed down a bit by the atmosphere. The equation for the kinetic energy is 0.5mv^2 so we can work out how big a meteor you would need..

KE = 0.5mv^2

m = 2KE/v^2
= 2*4.184*10^9/1,000^2
= 8,000kg

## 1. How are impact and detonation calculations used in scientific research?

Impact and detonation calculations are used in scientific research to study the effects of explosions and high-velocity impacts on structures and materials. These calculations help scientists understand the behavior of materials under extreme conditions and can be used to design safer structures and protective materials.

## 2. What factors are taken into account in impact and detonation calculations?

In impact and detonation calculations, factors such as the type and size of the explosion, the material properties of the target, and the distance between the explosion and the target are taken into account. Other factors such as temperature, pressure, and shock waves are also considered in these calculations.

## 3. What techniques are used to perform impact and detonation calculations?

There are various techniques used to perform impact and detonation calculations, including numerical simulations, analytical methods, and experimental testing. Numerical simulations involve using computer programs to model and simulate the impact or detonation event. Analytical methods involve solving equations and using mathematical models to predict the behavior of materials. Experimental testing involves conducting physical experiments to measure the effects of an explosion or impact on a material.

## 4. What are the applications of impact and detonation calculations?

Impact and detonation calculations have numerous applications in various fields such as aerospace, military, and civil engineering. They are used to design and test materials for aircraft, spacecraft, and other structures that may encounter extreme conditions. These calculations are also used in military research to study the effects of explosives and weapons. In civil engineering, they are used to design and test blast-resistant structures and protective barriers.

## 5. How accurate are impact and detonation calculations?

The accuracy of impact and detonation calculations depends on various factors such as the quality of input data, the complexity of the event being modeled, and the chosen calculation technique. Generally, these calculations have a high level of accuracy and can provide valuable insights into the behavior of materials under extreme conditions. However, it is important to validate these calculations with experimental data to ensure their accuracy.

• Classical Physics
Replies
12
Views
1K
• Mechanical Engineering
Replies
22
Views
1K
• Mechanics
Replies
6
Views
2K
• Mechanics
Replies
1
Views
8K
• Classical Physics
Replies
20
Views
1K
• Mechanics
Replies
2
Views
3K
• Mechanics
Replies
7
Views
10K
• Mechanics
Replies
2
Views
2K
• Mechanics
Replies
5
Views
80K
• Mechanics
Replies
6
Views
2K