Solar Insolation Calculator
Find the daily solar radiation on a horizontal surface for any latitude and day of the year, before the atmosphere takes its cut.
What it does
A solar panel does not see the sun straight above it, and it does not see it all day. The amount of radiation that lands on a flat roof or on a field is a geometry problem, and the answer is the single most useful number in solar energy work, because everything else is a correction on top of it.
The NASA technical note behind this page describes the calculation in three parts. First the total daily solar radiation arriving on a horizontal surface outside the atmosphere, second the zenith angle of the sun, which comes from the latitude, the declination and the hour angle, and third the sunrise and sunset hour angles. The same note points out that this quantity is needed for meteorological and climatological work.
The engineering text behind this page splits it into the pieces a calculator wants. The solar declination depends only on the day of the year and is 23.45 degrees at the solstice. The extraterrestrial irradiance on a surface square to the sun follows the solar constant of 1367 W per m2 with a small correction for the distance of the Earth from the Sun through the year. The daily radiation on a horizontal surface then follows from those two angles.
This page runs the whole calculation for a latitude and a day of the year, and returns the daily radiation, the average irradiance across the day, the equivalent peak sun hours, the irradiance square to the sun and the sunset hour angle. The numbers are the ones a designer would use to size a system before any correction for the atmosphere.
The limits are obvious and worth stating. This is radiation outside the atmosphere, so the real figure at the ground is lower by the clearness index of the site. The calculation ignores the angle of the surface, so it is for a horizontal one, and it ignores any shading, any reflection off snow and the fact that the sun is not a point.
How to use
- Choose metric or imperial for the radiation and the irradiance.
- Enter the latitude in degrees, with north positive.
- Enter the day of the year, where day 172 is close to the June solstice and day 355 close to the December one.
- Read the daily radiation, which is the energy that lands on one square metre of horizontal ground in a day.
- Read the average irradiance to compare with a nameplate power, and the peak sun hours to compare with the output of a reference test.
- Remember this is the ceiling. Multiply by the clearness index of the site to get a realistic figure.
Formula
- H
- Daily radiation on a horizontal surface, in joules per square metre per day.
- Io
- Extraterrestrial irradiance square to the sun, in watts per square metre.
- omega
- Sunset hour angle in radians, from the cosine of the angle between the sun and the horizon.
- declination
- Angle of the sun north or south of the equator, which depends on the day of the year.
- latitude
- Angle of the site north or south of the equator.
The declination comes from the day of the year as 23.45 degrees times the sine of 360 times 284 plus the day number, over 365. The extraterrestrial irradiance follows the solar constant with a correction of 0.034 times the cosine of 360 times the day number over 365.25, which is the small change in the distance of the Earth from the Sun. The sunset hour angle is the angle whose cosine is minus the tangent of the latitude times the tangent of the declination.
Worked example
Example: 40 degrees north on the June solstice
Defaults: metric units, a latitude of 40 degrees and day 172 of the year.
- The declination is 23.45 x sin(360 x 456 / 365), which comes to 23.45 degrees, the sign of a northern summer.
- The extraterrestrial irradiance is 1367 x (1 + 0.034 x cos(360 x 172 / 365.25)) = 1321.3 W per m2.
- The sunset hour angle has a cosine of minus tan(40) x tan(23.45) = -0.3638, so omega is 111.34 degrees, which is 1.9433 radians.
- Inside the brackets: 1.9433 x sin(40) x sin(23.45) = 0.4971, and cos(40) x cos(23.45) x sin(111.34) = 0.6542, so the total is 1.1513.
- The daily radiation is 86400 x 1321.3 / pi x 1.1513 = 41,849,000 J per m2, which is 41.849 MJ per m2 per day.
- Spread over 86400 s that is an average of 484.4 W per m2, and 41.849 MJ is 11.62 kWh per m2, so 11.62 peak sun hours.
- The tool shows 41.849 MJ per m2 per day, 484.4 W per m2, 11.62 h, 1,321.3 W per m2 and a sunset hour angle of 111.34 degrees.
Reference table
| Latitude | Daily radiation | Average irradiance | Peak sun hours |
|---|---|---|---|
| 0 degrees | 33.337 MJ per m2 per day | 385.8 W per m2 | 9.26 h |
| 20 degrees | 39.487 MJ per m2 per day | 457.0 W per m2 | 10.97 h |
| 40 degrees | 41.849 MJ per m2 per day | 484.4 W per m2 | 11.62 h |
| 60 degrees | 41.317 MJ per m2 per day | 478.2 W per m2 | 11.48 h |
Limits
- The number is radiation outside the atmosphere, so the energy that reaches the ground is lower by the clearness index of the site, which depends on cloud, haze, dust and altitude.
- It applies to a horizontal surface. A tilted or tracking surface collects a different amount, and so does a surface with any shading.
- The sun is treated as a point, so there is no allowance for the angle of the solar disc, and no correction for the changing distance of the Earth from the Sun beyond the simple term used here.
- It says nothing about how the radiation is split between the direct beam, the diffuse sky and the reflection from the ground.
Common mistakes
- Using this figure as the real output of a system. It is the ceiling, and a good site collects roughly half of it.
- Confusing the daily energy with the average power. The average irradiance of 484.4 W per m2 over a 24 hour day is the same energy as a much larger power over a short afternoon.
- Getting the sign of the latitude wrong. South is negative, and a southern latitude in June has a low sun.
- Using day 172 when you meant day 1, which changes the declination completely.
- Forgetting that the peak sun hours figure assumes a system that faces the sun and loses nothing, so it is not a panel rating.
Try it yourself
Quick checks. Answers sit tucked under each question.
How much radiation lands on a horizontal surface at the equator on the same day?
33.337 MJ per m2 per day, which is an average of 385.8 W per m2 over the whole day and 9.26 peak sun hours. The equator gets a lower daily total than 40 degrees north at this point of the year, because the days are short but the sun is never high, while the extra daylight further north more than makes up for the lower angle.
What does the average irradiance of 484.4 W per m2 mean for a panel?
It is the average power landing on every square metre of horizontal ground across the whole day, sun and night together. A panel that converts the radiation it receives at a typical efficiency of 20 percent would average about 97 W per m2 over the day, which is why the output of a real array is a good deal below the peak.
Why is the sunset hour angle useful on its own?
It is the angle at which the sun goes below the horizon, so it sets the length of the day. At 40 degrees north on day 172 it is 111.34 degrees, which is 14.85 hours of daylight, and the daily radiation follows from it together with the declination and the latitude.
Questions
What is insolation?
The energy that reaches a surface from the sun over a period of time, in joules or megajoules per square metre. The NASA note behind this page calls the daily total on a horizontal surface the quantity of interest for meteorological and climatological purposes.
What is the difference between insolation and irradiance?
Irradiance is the power per unit area at one moment, in watts per square metre, and insolation is the energy per unit area over a period, in megajoules per square metre. Multiplying the average irradiance by the length of a day gives the insolation.
What is peak sun hours?
The insolation expressed as the number of hours of full sunlight it is worth, so a site with 5 peak sun hours a day delivers the energy of 5 hours at 1 kW per m2. It is the usual way of stating a site's solar resource.
How does the solar constant enter the calculation?
It is the starting point, 1367 W per m2, the power that lands on a surface square to the sun outside the atmosphere. The engineering source behind this page notes that it has been estimated by several groups and that 1367 is the widely accepted value, with a small seasonal correction for the distance of the Earth from the Sun.
Why does the day of the year matter so much?
It sets the declination, which is the angle of the sun north or south of the equator, and that angle changes both the height of the sun and the length of the day. Between the solstices the declination swings by 47 degrees, which is far more than any other factor in the calculation.
What is the clearness index?
The ratio of the radiation that actually reaches the ground to the radiation outside the atmosphere. The engineering text behind this page uses it to describe how much of the available energy the atmosphere and the clouds let through.
Sources
- NASA technical report: daily solar radiation on a horizontal surface, with the zenith angle relations
- Solar energy fundamentals: declination, extraterrestrial irradiance and the daily radiation on a horizontal surface