Solar Panel

> *"Directly generate electricity using sunlight, and the actual generation capacity is related to the planet's solar energy ratio. Not working at night, use accumulators can store excess energy during the day."*
>
>

## Overview

**Solar Panels** are an early-tech renewable energy provider. However, they require High-Purity Silicon (which your starting planet lacks), generally need Accumulators to be viable (which are more technologically advanced and more complicated to build), and produce very little power for their footprint. As a result, they are often ignored in favor of continuing to burn Coal and/or Hydrogen in Thermal Power Stations. Unlike Wind Turbines, they do not have a power distribution area and can be placed immediately adjacent to each other.

## Mechanics

The formula for output "strength" is:

`currentStrength = planet.luminosity * clamp(2.5 * (sx*x + sy*y + sz*z) + 0.857244491577148)`

where clamp() clamps in the range 0-1, and planet.luminosity can be seen on the starmap in percent. (x, y, z) is the normalized position of the Solar Panel, and (sx, sy, sz) is the normalized sun vector. All percentage values are multiplied by 100 for display. The actual power output will be 360kW \* currentStrength.

The vector dot-product means that the panel will keep producing nearly-full power at the moment the sun has set, but quickly fall off from there. It also means that placing Solar Panels at the poles would be optimal (assuming no axial tilt), because you would get a constant 85.72% output - except those positions are better saved for Ray Receivers, for similar reasons.

### Detailed Math

For the general case of non-zero axial tilt everything gets very complicated - you potentially have to deal with the eccentricity of the orbit, the orbit length in (solar) days, etc. But with zero tilt those complications go away, and we can simplify the vector dot-product down to `cos(𝜃)cos(𝜑)`, where 𝜃 is the rotation of the planet and 𝜑 is the latitude of the Solar Panel (0 at the equator, 𝜋/2 at the pole(s).)

This means the overall daily irradiation (as a fraction of full illumination) is:

1
π

∫

0

π

clamp
⁡

(

2.5
cos
⁡
(
θ
)
cos
⁡
(
φ
)
0.857244491577148

)

d
θ

{\displaystyle \frac{1}{\pi} \int\_{0}^{\pi} \operatorname{clamp}\left(2.5 \cos(\theta)\cos(\varphi) + 0.857244491577148 \right) \, d\theta}

The clamp function makes this awkward to solve in closed form. It breaks down into three cases, corresponding to the behavior of the clamp:

* 𝜑 > 1.51366305, corresponding to angles over 86.73 degrees. At these high latitudes, the clamp never saturates and so the integration is straightforward:

1
π

∫

0

π

clamp
⁡

(

2.5
cos
⁡
(
θ
)
cos
⁡
(
φ
)
0.857244491577148

)

d
# θ

1
π

∫

0

π

2.5
cos
⁡
(
θ
)
cos
⁡
(
φ
)
0.857244491577148

d
# θ

1
π

(

−
2.5
sin
⁡
(
θ
)
cos
⁡
(
φ
)
0.857244491577148

θ

)

0

π

=

0.857244491577148

{\displaystyle \begin{align}
&\frac{1}{\pi} \int\_{0}^{\pi} \operatorname{clamp}\left(2.5 \cos(\theta)\cos(\varphi) + 0.857244491577148 \right) \, d\theta =\\
&\frac{1}{\pi} \int\_{0}^{\pi} 2.5 \cos(\theta)\cos(\varphi) + 0.857244491577148 \, d\theta =\\
&\frac{1}{\pi} \left. \Big(-2.5 \sin(\theta)\cos(\varphi) + 0.857244491577148 \, \theta \Big)\right|\_{0}^{\pi} =\\
& 0.857244491577148
\end{align}}
* 1.22079634 < 𝜑 < 1.51366305, corresponding to angles between 69.95 and 86.73 degrees latitude. In this range, the clamp saturates on the high end but never on the low end. The detailed math is omitted because it's a simplified version of:
* 𝜑 < 1.22079634, corresponding to angles less than 69.95 degrees latitude. The majority of the planet's surface is also the most complicated to model; the clamp saturates at both ends and has to be dealt with specially.

Let
# k
0.857244491577148

1
π

∫

0

π

clamp
⁡

(

2.5
cos
⁡
(
θ
)
cos
⁡
(
φ
)
k

)

d
# θ

1
π

(

∫

0

cos

−
1

⁡

(

1
−
k

2.5
cos
⁡
(
φ
)

)

1

d
θ

∫

cos

−
1

⁡

(

1
−
k

2.5
cos
⁡
(
φ
)

)

cos

−
1

⁡

(

−
k

2.5
cos
⁡
(
φ
)

)

2.5
cos
⁡
(
θ
)
cos
⁡
(
φ
)
k

d
θ

)

=

1
π

(

c
o

s

−
1

(

1
−
k

2.5
cos
⁡
(
φ
)

)

(
−
2.5
sin
⁡
(
θ
)
cos
⁡
(
φ
)
k

θ
)

cos

−
1

⁡

(

1
−
k

2.5
cos
⁡
(
φ
)

)

cos

−
1

⁡

(

−
k

2.5
cos
⁡
(
φ
)

)

)

=

1
π

(

(
1
−
k
)
c
o

s

−
1

(

1
−
k

2.5
cos
⁡
(
φ
)

)

k

c
o

s

−
1

(

−
k

2.5
cos
⁡
(
φ
)

)

−
2.5

1

(

−
k

2.5
cos
⁡
(
φ
)

)

2

cos
⁡
(
φ
)
2.5

1

(

1
−
k

2.5
cos
⁡
(
φ
)

)

2

cos
⁡
(
φ
)

)

=

1
π

(

(
1
−
k
)
c
o

s

−
1

(

1
−
k

2.5
cos
⁡
(
φ
)

)

k

c
o

s

−
1

(

−
k

2.5
cos
⁡
(
φ
)

)

−

2.5
2

cos
⁡
(
φ

)
2

k
2

2.5
2

cos
⁡
(
φ

)
2

(
1
−
k

)
2

)

{\displaystyle \begin{align}
&\text{Let }k=0.857244491577148\\
&\frac{1}{\pi} \int\_{0}^{\pi} \operatorname{clamp}\left(2.5 \cos(\theta)\cos(\varphi) + k \right) \, d\theta =\\

&\frac{1}{\pi} \biggl( \int\_{0}^{\cos^{-1}\left(\frac{1-k}{2.5\cos(\varphi)}\right)} 1 \, d\theta +
\int\_{\cos^{-1}\left(\frac{1-k}{2.5\cos(\varphi)}\right)}^{\cos^{-1}\left(\frac{-k}{2.5\cos(\varphi)}\right)} 2.5 \cos(\theta)\cos(\varphi) + k \, d\theta \biggl) =\\

&\frac{1}{\pi} \biggl( cos^{-1}\left(\frac{1-k}{2.5\cos(\varphi)}\right) +
(-2.5 \sin(\theta)\cos(\varphi) + k \, \theta) \Biggl|\_{\cos^{-1}\left(\frac{1-k}{2.5\cos(\varphi)}\right)}^{\cos^{-1}\left(\frac{-k}{2.5\cos(\varphi)}\right)} \biggl) =\\

&\frac{1}{\pi} \biggl( (1-k) cos^{-1}\left(\frac{1-k}{2.5\cos(\varphi)}\right) +
k\, cos^{-1}\left(\frac{-k}{2.5\cos(\varphi)}\right) +\\
&-2.5 \sqrt{1+\left(\frac{-k}{2.5\cos(\varphi)}\right)^2} \cos(\varphi) +
2.5 \sqrt{1+\left(\frac{1-k}{2.5\cos(\varphi)}\right)^2} \cos(\varphi) \biggl) =\\

&\frac{1}{\pi} \biggl( (1-k) cos^{-1}\left(\frac{1-k}{2.5\cos(\varphi)}\right) +
k\, cos^{-1}\left(\frac{-k}{2.5\cos(\varphi)}\right) -\\
&\sqrt{2.5^2\, \cos(\varphi)^2+k^2} +
\sqrt{2.5^2\, \cos(\varphi)^2+(1-k)^2} \biggl)
\end{align}
}

Here is a link to a Google plot of the resulting function in the relevant domain: Interactive Graph. The values range from 54.872% at the equator to 81.138% at the 69.95 degree cutover, where they increase only slightly to the maximum of 85.724% around the poles.

### Tips & Tricks

* You can place your solar panels in a ring around the equator of a planet, if you want a consistent power supply without using any accumulators.

* You can place your solar panels on and near both poles of a planet. This has two advantages compared to placing them near/around the equator: 1) each panel typically produces 30-80% more energy per planetary day because they will always be getting sunlight, even when the sun is at its farthest away point, and 2) it is possible to get a consistent supply of power with as few panels as desired, compared to placing them around the equator, which forces the player to evenly place panels around the entire equator. It should be noted that, the higher the axial inclination of the planet, the more energy a panel near the pole of the "Summer Zone" will produce per day compared to a panel near the pole of the "Winter Zone". However, the further away the panel is from the pole, the less this effect is visible.

## Crafting Recipes

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### Crafting Into

Solar Panel

16s10105

### Total Raw Materials

512.520

Crafting Energy (not including mining): **17.8 MJ**