Compression Blast

 

A compression blast or and 'air blast' is a very cool mechanic built into this game, which many other games do not have. A compression blast comes from the pyro's primary weapon called the flamethrower. It pressurizes air inside the tank and then uses it to blast out fast shooting air in less than a second. Doing so creates a huge amount of kinetic energy from the air. In the video game, compression blasts can deflect most projectiles, push enemies into the air, and extinguish friendly teammates from fire. The way the flamethrower works is that there is an intake opening on the bottom of the flamethrower which air is compressed in the tank and shot out from the top. 

 

The air blast happens so fast and that is why this weapon can create huge amounts of energy. Like before flamethrowers can deflect almost all projectiles: pipes, sticky bombs, arrows, baseballs, flares, and many more. In fact it reflect more than just one rocket, it can reflect as many rockets. Doing so would result in the flamethrower producing an infinite amount of energy. 

 

It takes just about 0.36 s - 0.48 s to have any object change direction as well as conserve the same amount of kinetic energy. 

 

 

Rockets

 

For this calculation, I am only going to do calculations on one reflected rocket

 

First we will need to calculate the acceleration of the rocket. 

Because the flamethrower follows the conservation of energy, this means the final velocity will be exactly the same as the initial velocity, by reversed. It also follows the conservation of momentum which means that the momentum in to the airblast is conserve and is exactly the same as the out. 

 

vi = 21 m/s

vf = -21m/s

t = 0.42 s

 

a = vf - vi 

          t

a = 21m/s - (- 21 m/s)

                 0.42 s

a = 100 m/s/s

 

This means that the flamethrower accelerates the rocket at 100 m/s/s in the opposite direction, which is fast!

 

We can then use the forces equation: Fnet = ma 

Since there is only one force acting on the projectile since it is not affected by gravity, this means that: 

Fnet = Fa

 

m = 2.6 kg 

a = 100 m/s/s

 

Fnet = 2.6 kg x 100 m/s/s

Fnet = 260 N 

 

We can use the Fa to convert to kg to see how much 260 N feels like.

F = ma

260 N = m x 9.8 m/s/s

m = 26.5 kg

 

This means that 260 N is like lifting 26.5 kg which seems like such little force used considering it is reflecting a rocket.

 

The flamethrower follows the law of conservation of energy, where the energy going towards the flamethrower is the same energy from the reflected object going out. The only way that is possible for the flamethrower, is if the compression blast puts in double the amount of energy than the rocket. This would mean it would throw out 570 J x 2 = 1140 J of energy

 

PSI

 

To calculate the pressure it takes to reflect this projectile, we would need to find the surface area of it. The PG-7VL has a surface area of a cone with a diameter of 93 mm. From tracker we can now find the surface area since we have found the height of the cone shape at the tip of the war head.

 

 

height from tracker = 147 mm

d = 2 x r

r = 93/2 mm

r = 47 mm

 

 

area of cone = 3.14 x r (r + sqrt (h^2 + r^2))

area = 3.14 x 47 mm ( 47 mm + sqrt (147 mm^2 + 47 mm ^2))

area = 147.6 mm x (47 mm + 154 mm)

area =  147.6 mm x 201 mm

area = 29,668 mm^2

 

 

Now we can find how much newton of force is applied over a certain area:

 

     260 N      = 0.00876 N/mm^2

29,668 mm^2

 

Now with a simple conversion to pounds per square inch and then we can determine the PSI.

0.00876  N/mm^2  =  1.27 PSI This is also not that much pressure as it is about the pressure of the heart less than 60 BPM. However, If you were to calculate how much pounds you would feel we would get around 53.4 lbs of force. 1.27 PSI does not seem to be a lot of pressure, but in the video game world, that is all is needed to reflect a rocket traveling at 21 m/s

 

Enemies

 

For enemies, the energy and forces also change as well. I calculated the forces that would be acting on a demomen. From tracker, I can see that the air blast accelerates the demoman from 280 HU/s (5.33 m/s) to the left to 18.33 m/s in the opposite direction upwards in just around 0.209 s. This gives an acceleration of:

 

a = vf - vi

           t

a = -18.33 m/s - 5.33 m/s

                   0.209 s

a = - 23.66 m/s

            0.209 s

a = 113 m/s/s

 

We can use Fnet = ma to find the net force of the object being accelerated.

Mass of demoman is 90.7 kg found on the gravity page. 

 

 

m =90.7 kg

a = 113 m/s/s

g = 15.24 m/s/s

Theta = 42.6

 

Fnet = 90.7 kg x 113 m/s/s

Fnet = 10,249.1 N = 10,200 N 

 

Fg = 90.7 kg x 15.24 m/s/s

Fg = 1383 N = 1380 N

 

This force is equivalent to:

 

10,200 N = m x 9.8 m/s/s

m = 1,040 kg on earth

 

 

PSI

 

Now we can calculate exactly how much pressure we need to lift a person. The body surface area of an adult man is 1.9 m^2.

1.9m^s / 2 = 0.95 m^2 which is how much area is affected by the air pushing the person upwards.

This means that we would need 10,200 N of force over 0.95 m to life a person in TF2.

 

10,200 N   = 10736 N/m^2

  0.95 m^2

 

This can now easily be converted into PSI, the more common unit for measuring pressure. 

10736 N/m^2 also is only about 1.56 PSI (pounds per square inch) That is not that much pressure (about the same as heart rate 80 BPM)  However we also have to consider the amount of pressure that gets lost from the flamethrower since not all the air particles exactly hit the person. In this case, the end of the flamethrower is very wide which leaves room for a lot of the energy pressure to escape. Therefore we can only assume that 1.56 PSI is actually hitting the demoman and that the flame thrower actually puts out more than 1.56 PSI

 

Arrows

 

 

Unlike the other 2 projectiles mentioned, the arrow projectile has a much smaller surface area. Not to mention that it has a higher velocity than both the other projectiles. Therefore we can also assume that the PSI that must be applied on the arrow in order to reflect it will be significantly higher. Following the laws of energy and momentum we can find the acceleration of the arrow.

 

Now there are three different speeds on the arrow from slowest to fastest: 34.3 m/s, 45.72 m/s and 49.5 m/s.

The one that will require the most pressure to reflect will be the fastest, so there will only be one calculation.

 

It takes about 0.209 seconds for the arrow to accelerate at full speed.

There are two ways we can find the force applied on the arrow.

 

Energy

 

W = Fd 

F = W/d

 

or 

 

Forces

 

a = vf-vi

         t

a = 49.5 m/s

       0.209 s

a = 237 m/s/s

 

Fnet = ma

Fnet = Fa

Fa = ma

Fa = 0.0697 kg x 237 m/s/s

Fa = 16.5 N 

 

The percent deviation is 17.3 N -16.5 N   x 100 %  = 4.8%

                                                      16.5 N

This is a pretty accurate calculation with a low percent deviation. However we shall stick  to using the force equation. 

 

Now we have to find the surface area of the arrow

 

Since pressure is force/area

We can easily find the PSI required to reflect an arrow 

 

16.5 N /     =   lb/in^2

 

Uncertainties 

 

There are a few uncertainties to point out. Firstly for projectiles which is everything but the rocket, the velocity changes over time, as they are all subjected to gravity. This will directly affect the pressure needed to reflect the projectile. It would be easier to reflect a pipe bomb right the the top of its tragectory than it is at the bottom. So for the calculations above I just use the intial velocities as the reference. There is also uncertianties in the tracker, although it is pretty accurate it make mess up the data. This is because projectiles have such a very, very fast acceleration once it launches that one pixel may affect a lot of the acceleration. For the surface areas, there is also uncertainties. I only took the average body surface area of a male and the rocket surface area may be a bit off considering there is a silver tip on the rocket which I did no account for.