**Calculations**

**In this section, the calculations are only approximate**, not exact truths;-) The calculations are presented so that an understanding of high school level physics is sufficient to verify their basis. **The main problem in these calculations is the lack of knowledge considering the efficiency of the EMU is recovering energy from the hurricane.** The EMU´s rotation on the water surface is impeded by the airfoils and by the moving of water from the front of the device to behind it. Therefore, the device should be further developed with the aim of minimizing the time that it actually moves in the water. In addition, **the most advantageous airfoil structure still needs to be determined.** Here the input of technical institutions or marine industry research units would be valuable. Another aspect needing further research is to come up with a suitable substance for storing third energy, and developing a way to produce this substance. If we are able to produce the familiar substances such as gasoline, diesel oil, and kerosene synthetically and efficiently enough, this should do.

**The amount of energy in a hurricane**

Let us assume a hurricane with a **diameter of 300 km**, average **wind speed of 50 m/s** and a **total air layer thickness of 8 km**.

The volume of moving air is approx. 5.7 e+5 cubic kilometers and the mass 7.3 e+11 tons. The momentary kinetic energy of the hurricane winds is approx, 9,2 e+17 J or **260 TWh. That is enough to replace the output of a nuclear plant for 32 years** (Lovisa). That energy will be enough for Finland for five years.

**Energy coming in from the sun to the Earth**

Energy from the sun enters the Earth’s atmosphere** 2 cal/cm2 /min**. The atmosphere **reflects approx. 35% energy back into space**.

This energy arrives to the surface of the Earth (the day side and a circle with a **diameter of 13,000 km**) with power of **120,000 TW**. This means that at noon, the energy arriving from the sun on one square kilometer on the equator contains **0.91 GW** of power, which is **equal to the output of the of the nuclear power plant (Lovisa in Finland).**

In 2014 the world energy consumption was **20 TW**, which is **0.17 ‰ of the energy arriving from the sun** to the surface of the Earth. This amount of power arrives from the sun at noon on the area of 22,000 km2 on the equator, i.e. a **squares with a side length of 150 km**.

Because the sun does not shine on the Earth´s surface all the time (night), this would mean roughly that **in Sahara all the energy needed in the world arrives on the square with a side length of approx. 300 km.**

**Information on processing carbon dioxide **

When carbon dioxide is collected of storage, the best is to condence it into a liquid form. This requires less than **1 MJ/kg** of energy (0.6 MJ/kg at condensation point -57 ° C), in other words only a few percent of the total amount of energy we get from fuel.

**T****he thermal energy released during the condensation need not go to waste, **but could be used for heating houses and water, for example. Let us imagine that people produce all the energy (**20 TW**) by burning carbon compounds and 3% of this energy is used for liquefying carbon dioxide. The **thermal energy thus produced would amount 600 GW, which means that the yearly production of the thermal energy would be 5,300 TWh in a year**. For the sake comparison, in 2013 the energy consumed for heating by Finnish households was** 55 TWh**.

If all the energy that people use (**20 TW**) were to be produced by burning carbon compounds, the energy is produced by burning **gasoline 450 tn** or **wood 1300 tn** in every second. **This is equal to burning 7.3 liter of oil or 17 kg of wood per day per person.** It is also interesting to know that the cube of air with a side length of 100 m contains approx. one ton of carbon dioxide.

**Theoretical production rates**

Taking into consideration the amount of energy contained in different fuels, we can calculate the production rates, assuming that the output of the EMU device producing these fuels with power of **20 Mw (efficiency of 100%)**.

**Methanol:** Amount produced 1.0 kg/s (1.3 l), or **86 ton/day,** carbon dioxide needed for production ** 120 ton/day. **

**Ethanol:** Amount produced 0.7 kg/s (0.9 l), or **57 ton/day**, carbon dioxide needed for production **110 ton/day**.

**Gasoline:** Amount produced 0.45 kg/s (0.6 l), or **39 ton/day,** carbon dioxide needed for production **120 ton/day.**

**Investment calculations 😉**

**Let us assume that the manufacturing costs of a 40 MW device are an estimated EUR 20 million and that the price of a 1.5 GW nuclear plant was approx. EUR 5,000 million.**

- The world´s energy consumption is currently at
**20 TW**. - This energy can be produced with
**13,000 nuclear power plants, the total price EUR 67 trillion**or

**with 2 000 000 EMU devices, the total price of which is EUR 40 trillion**assuming that the efficiency is 25%. - The maintenance costs could be assumed to be equal.
- The energy derived from hurricanes is virtually pollution-free.
- Additionally, the reducing effect on the destructive force of hurricanes could be significant.
- The total amount of US public debt is currently approx.
**Eur 17 trillion**.

In sum: A**ll of the world’s (pollution-creating) energy could be replaced with an investment equaling 2 – 4 times the amount of the US public debt.** – by the EMU devices and by solar energy.

**EMU energy yield calculations**

The maximum output of the electric generators can be influenced be the eccentric counterweights of the main shaft and by their positioning. The Earth’s gravity pulls the counterweight, which thereby tends to reach its bottom position. As described before, the **EMU’s efficiency decreases as displacement increases.**

Example: Let us imagine **an EMU device with the hull length of 40 m and the diameter of 12 m, with the airfoils height at 2 * 2.7 m,** then the surface area of the ellipsoid would be maximum** 5.5 acre**.

When the wind pushes the divice on the water, the wind and water (especially below the water surface) hittig the airfoils rotate it, controlled by computers.

**By calculating the dynamic thrust of the wind on the device **that can be utilized as such to rotate the device,** we can calculate the device´s power of rotation.** A part of this power is obviously lost in the whirlpools of water and wind.

**When the wind speed is at 40 m/s**, the dynamic pressure (0.5 * density * v*v) on the surface of device is approx. **10 mbar**. This pressure **pushes the device at power of 40 ton (ton=10 kN)**, assuming that 70% of the device is above the water surface.

The power input could be calculated using the formula **P = Thrust * propagation speed of the device**.

If the **peripheral speed of the rotating device is one-third of the wind speed**, the propagation speed of the device´s center of gravity would be approximately the same as the peripheral speed. **The efficiency of the wind thrust would be 5 Mw**, which would be the maximum power output of the generators of the device in that moment.

If the **wind speed were 20 m/s**, the dynamic pressure on the surface of the device is about **3 mbar**, which would push the device with force of** 10 ton** (ton=10 kN). If the peripheral speed of the rotating device is still one-third of the wind speed, the storm pushes the devece **with power of 1.2 MW**, so the production of energy is **less than 1.2 Mw.**

**What the energy output would be in reality is a small mystery.** It could, however, be figured out through careful experiments with prototypes in technical institutions or in marine industry research units.

The size of the device need not be equal to the one presented above. In similar devices the theoretical power output will change exponentially in the second power. The power output of a **160 m** long device would then be **80 Mw and 20 Mw** repectively, assuming that the wind speeds are the same. On the other hand, the power output of the EMU will change exponentially in the second power of wind speed, too. ie When the speed of the wind doubled, the energy output of the device is quadrupled.

The power output of the generators in EMU would be** calculated also in another way** after the following example.

Let us assume that the device described in the previous example has a** 40 Mg** (40 tons) counterweight, and that the distance between the center of gravity of this counterweight and the axis of rotation is **r** (5.0m). The force with which the gravity pulls the counterweight is **400 kN**. When the wind rotates the device, the counterweight will also rotate with the device controlled by the computers, affecting the moment of the force resisting the rotation of the device. The force F of the associating bending moment of the force would be at the maximum 400 kN, when it resides at the high of the main shaft. The power of moment of the force F can be calculated **P = F * r * angular velocity**, which is the maximum power output at this moment. But the **r * angular velocity = v, **which is in princible the peripheral velocity of the centre of gravity of counterweight with respect to the main shaft (as the matter of fact the centre of gravity of the counterweights remains at the place and the outermost shell of the EMU rotates).

If the “peripheral speed of counterweight´s center of gravity” is **13 m/s **(third part of the above wind speed)**,** the energy output of the EMU´s generators could be **400kN x 13 m/s = 5 Mw**.

**Cost estimations**

**This cost estimation is for an EMU device that could produce methanol at the output of 10 MW.**

When the efficiency of the device will be 25%, the EMU would have to operate at an output of** 40 MW**.

**The sail area needed to create this kind of output would have to be 4 000 m2, with the wind speed at 150 km/h (= n. 40m/s).**

Manufacturing the hull of the said device ( length 110m) could be expected to cost around **Eur 20 million**, and the investment costs relating to the methanol production equipment, the storage facilities and the automatic steering could be estimated at **EUR 5 million**. When financed with straight-line payments on a 10 years loan, the **annual costs would be EUR 3.5 million.**

Additionally, the** ongoing expenses for maintenance, repairs, transfers etc. could be estimated EUR 5,000 per day, amounting to EUR 1.8 million per year.**

Based on the above, the annual costs of an EMU of the said type would be approximately **EUR 5.3 million**, excluding methanol transportation costs and taxes.

When producing methanol at an effective output of 1**0 MW**, the EMUs would produce 43 tons, or 55 m3 of methanol in a day.

With the global market price of methanol at **370 eur per ton**, the **value of methanol produced daily would be about EUR 16,000**, when the average wind speed is 150 km/h.

With annual costs of the device estimated at EUR 5.3 million, **the production of 330 days in a year would be enough to create a profit.**

### When comparing these calculations with oil production, the following can be stated:

The energy content of methanol is approx. **45%** of that of hydrocarbons. In terms of energy content, the** daily methanol output of 43 tons would thus be equal to 160 oil barrels**. With the price of a crude oil barrel at **USD 75**, the** total price of oil equaling the methanol would be USD 12,000 or EUR 10,000.**

The proposed energy production method is also carbon neutral, which means that no CO2 emission charges would need to be paid with regard to the energy.

In addition, what could also be factored in on the cost calculations are the subsidies which the coastal states ravaged by hurricanes will pay for a solution to their problem.

Considering that the oil in this example is unprocessed crude oil, **use of the EMU devices seems quite competitive compared to hydrocarbons**, the resources of which are finite.

**Asymmetric moving of EMU devices**

In the image, each blue line represents one EMU around the hurricane eye. Here the fleet of the EMUs derives energy **asymmetrically** from the strong winds around the edges of the hurricane eye. The animation shows the movement of the hurricane eye for a moment, paises for a few seconds and then starts moving again.

**Here the EMUs are concentrated to travel along one edge of the hurricane eye and operate there at maximum output.**

The arched arrows indicate how the direction of the wind tends to change when the wind hits EMUs diagonally. Particularly interesting is that the winds affecting on different sides of the hurricane eye tend to turn to the same direction (in this image, to the right). This means that changes in wind direction subject the air mass to a force that continuously affects to the same direction.

**A rough calculation of the force that continuously affects the position of the hurricane eye** can be made as follows: with the wind blowing at 150 km/h, the device’s sail areas will be subject to a pressure of approx. 0.01 atm, which is a force of **10 tons per acre**.

In terms of vector analysis, if the device’s axis of rotation forms a 30-degree angle with respect to the wind, then **50% of the wind force (scalar) turns to the direction of the device’s axis,** while 87% of the wind force hits the surface of the device perpendicularly and actually rotates it. It is thus reasonable to assume that **the wind that has hit the device and been reflected off it will push the air in front of it at a force of 5 tons per one acre of sail area**. With the sail areas of the device varying between 2-50 acres and the number of devices easily in the hundreds, it is easy to understand that **this continuously affecting force of many kilotons** could well contribute towards a change in the hurricane eye’s position.**See also another explanation .**