Rip Apart my Calculation

[Yogi]

I'm trying to calculate the energy output of a mecha called "The Shaft". It is from the Anime, Geneshaft. I would like to make sure my calculations are correct.

I am basing my calculations on the fact that The Shaft, after it takes it's reactor out of itself and sends it in a ring around Europa, is able to act like a second sun.

The sun outputs 3.86e25 watts of energy and the Earth is 1.5e12 meters from the Sun. Therefore, at this point the energy is spread out over 4*pi*(1.5e12)^2 or 1.83e25 square meters, or 0.732 watts per meter square.

Now, some parts of the Earth get more energy than others. High Noon at the equator gets the most, while the poles and the sunset-sunrise areas get less. However, since the energy is spread in a ring around Europa, then it's essencially "High Noon" everywhere. I will then take the "longest" part of the shadow cast by the Earth, and spread it to all point to the left and right as if they were both at the "high noon" point so that an effective r^2 square meter shadow is cast. This "energy shadow" is then divided by the actual surface area, which is 2*pi*r^2 (half of a sphere). With this, we can conclude that the average energy on earth at high noon is 0.1166 watts per square meter.

Europa has a raduis of 1.57e6 meters. That gives it a surface area of 1.973e13 square meters. Times 0.1166 watts per square meter, that gives us 2.3e12 watts of power. However, since the energy is radiated in a ring, at least half of it will be radiating AWAY from the planet. Therefore, the total power output of the Shaft is at least 4.6e12 or 4.6 Terrawatts of power.

[Darth Wong]

The sun outputs 3.86e25 watts of energy and the Earth is 1.5e12 meters from the Sun. Therefore, at this point the energy is spread out over 4*pi*(1.5e12)^2 or 1.83e25 square meters, or 0.732 watts per meter square.

Incorrect. The sun outputs 3.827E26 W, and the Earth is 150 million km, or 1.5E11 metres from the Sun. Both of your figures are out by one order of magnitude. At our range, the intensity of solar radiation is roughly 1350 W/m^2.

Now, some parts of the Earth get more energy than others. High Noon at the equator gets the most, while the poles and the sunset-sunrise areas get less. However, since the energy is spread in a ring around Europa, then it's essencially "High Noon" everywhere. I will then take the "longest" part of the shadow cast by the Earth, and spread it to all point to the left and right as if they were both at the "high noon" point so that an effective r^2 square meter shadow is cast. This "energy shadow" is then divided by the actual surface area, which is 2*pi*r^2 (half of a sphere). With this, we can conclude that the average energy on earth at high noon is 0.1166 watts per square meter.

That's much too complicated (and the resulting number is wrong anyway, even if the input number is correct). The surface area of one hemisphere is 2*pi*r^2, and the projected area is pi*r^2. Therefore, the surface area is twice as large as the projected area, and the average surface intensity on the side facing the sun is roughly half of the normal intensity.

Europa has a raduis of 1.57e6 meters. That gives it a surface area of 1.973e13 square meters. Times 0.1166 watts per square meter, that gives us 2.3e12 watts of power. However, since the energy is radiated in a ring, at least half of it will be radiating AWAY from the planet. Therefore, the total power output of the Shaft is at least 4.6e12 or 4.6 Terrawatts of power.

I'm not sure what this shaft is, but you should fix your numbers and then try it again.

[Yogi]

That's much too complicated (and the resulting number is wrong anyway, even if the input number is correct). The surface area of one hemisphere is 2*pi*r^2, and the projected area is pi*r^2. Therefore, the surface area is twice as large as the projected area, and the average surface intensity on the side facing the sun is roughly half of the normal intensity.

However, that is considering that some parts are at mornings, some afternoon, and some evening. In Europa, ALL of the planet is at "High Noon" so I need to calcualte the energy density for the planet at high noon. Any idea how I could do that?

[Darth Wong]

That's much too complicated (and the resulting number is wrong anyway, even if the input number is correct). The surface area of one hemisphere is 2*pi*r^2, and the projected area is pi*r^2. Therefore, the surface area is twice as large as the projected area, and the average surface intensity on the side facing the sun is roughly half of the normal intensity.

However, that is considering that some parts are at mornings, some afternoon, and some evening. In Europa, ALL of the planet is at "High Noon" so I need to calcualte the energy density for the planet at high noon. Any idea how I could do that?

That's easy. You need the location of the ring so that you can trig out the percentage of the planet's surface area that has a line of sight. Then, you simply perform some algebra and you're done.