The whole calculation in one line
Array kW = daily kWh ÷ (peak sun hours × derate factor)
That is it. Twenty kilowatt-hours a day, 4.5 peak sun hours, a derate factor of 0.80 for 20 percent system losses: 20 divided by 3.6 is 5.556 kW. Every quote you receive is this arithmetic wearing a suit. The reason quotes differ by 40 percent for the same house is not that anyone is doing different maths, it is that they are using different peak sun hours and different loss assumptions, and neither of those is usually printed where you can see it.
Panel count is the array size divided by the panel rating, rounded up, because you cannot buy two thirds of a module. Round up and the array is slightly larger than the calculation asked for, which is the direction you want to be wrong in.
Peak sun hours: the number that decides everything
A peak sun hour is one hour of irradiance at 1,000 watts per square metre, which is the condition panels are rated at. Saying a location gets 4.5 peak sun hours means the day's total solar energy on your array plane equals 4.5 hours at rating. It is not the number of hours of daylight and it is not the number of hours the sun is above the roofline.
This figure is location-specific, orientation-specific and season-specific, and it is the single largest source of disagreement between two estimates of the same system. The same array in the same city can vary by 25 percent between a south-facing 30-degree plane and a west-facing shallow one. Get your figure from a solar resource dataset queried for your coordinates, your tilt and your azimuth. Public irradiance databases exist for most of the world and give monthly figures as well as annual ones, which is what you want, because the monthly spread is what determines whether the system works in winter.
What moves it:
| Factor | Effect on annual yield |
|---|---|
| Latitude and cloud climate | The dominant term. Sunny high-desert sites roughly double the annual insolation of cloudy maritime ones. |
| Tilt away from optimum | Modest. Anything within about 15 degrees of the optimum tilt costs only a few percent. |
| Azimuth away from the equator-facing direction | Grows quickly past about 45 degrees off. East and west planes commonly give 10 to 20 percent less than south in the northern hemisphere. |
| Shading | Non-linear and brutal. A single shaded module can drag a whole string down unless the array uses optimisers or microinverters. |
What the 20 percent of losses is made of
The derate factor covers everything between the sunlight hitting the glass and the energy arriving where you count it. Nobody has one number; they have a stack of small ones multiplied together.
| Loss | Typical share | Why |
|---|---|---|
| Module temperature | 5-12% | Panels are rated at 25 °C cell temperature and a roof-mounted module in summer runs far hotter. Output falls a few tenths of a percent per degree. |
| Inverter conversion | 2-4% | Modern inverters peak around 97-98% and do worse at very low load. |
| Soiling | 1-5% | Dust, pollen, salt, bird mess. Worse in dry climates with long gaps between rain. |
| DC and AC wiring | 2-3% | Ordinary resistive loss over the run, which is why conductor sizing matters on long roof-to-inverter runs. |
| Mismatch and tolerance | 1-3% | Modules in a string are not identical and the string follows the weakest. |
| Shading, snow, downtime | Site-specific | Can be zero or can dominate everything else. |
Twenty percent total is a defensible planning default. Fourteen is achievable on a cool, clean, unshaded, well-ventilated array. Twenty-five or more is what you should use if the array sits on a hot dark roof in a dusty place, or if anything shades it in the morning.
Roof area, and the area you actually have
Panel area is easy: a common 400 watt module is roughly 74 by 45 inches, which is about 23 square feet, so the array is 23 square feet per panel plus whatever you allow for spacing. The allowance is the part people underestimate. Pitched roofs need clearance at ridges and eaves and around penetrations, and many jurisdictions expect access pathways for fire service — the specific dimensions differ by adopted code and by whether the roof has other means of access, so ask locally rather than assuming. Flat roofs on ballasted racking need row spacing to stop each row shading the next, and at higher latitudes that spacing can consume as much area as the panels themselves.
The 25 percent default here is for a straightforward pitched roof. On a flat roof with tilted racking, 60 to 100 percent extra is more realistic.
What this sizing does not settle
An array size is a shopping figure, not a design. String lengths have to fall inside the inverter's voltage window with cold-morning open-circuit voltage accounted for, which is what actually determines how panels group together. Roof structure has to carry the dead load and the wind uplift, and older roofs sometimes cannot. And the array is only worth building on a roof with life left in it, because taking a system off and putting it back to replace shingles costs real money. Anything that ties a solar system to a building supply or to the grid is permit and inspection territory in most places. Interconnection agreements, rapid shutdown provisions, labelling, disconnect placement and who is allowed to do the work vary by jurisdiction and by utility, and the code edition your Authority Having Jurisdiction has adopted is what governs the installation, not a web page. Use these numbers to plan and to price, then have the design reviewed by someone who knows what your utility and your inspector expect.
Questions people ask
How many solar panels do I need for a house?
Work it from consumption rather than from house size, because two identical houses can differ threefold. Take twelve months of bills, add the kilowatt-hours, divide by 365 for a daily average, and put that in the field above with your own peak sun hours. As orientation only: a household using 20 kWh a day at 4.5 peak sun hours and 20 percent losses lands on about 5.6 kW, which is fourteen 400 watt panels and roughly 400 square feet of roof once you allow for setbacks. Move the peak sun hours to 3.5 and the same house needs eighteen panels. That is why nobody can answer this question without knowing where you live.
Should I size the array for the annual average or for winter?
It depends entirely on what absorbs the shortfall. A grid-tied system can be sized on the annual average, because the grid banks your summer surplus and hands it back in winter, subject to whatever your utility actually pays or credits for exports. An off-grid system has no such bank, so it has to be sized on the worst month, and at mid latitudes that commonly means one and a half to two and a half times the array an annual-average calculation suggests. The middle path most off-grid installations take is to size somewhere between the two and cover the gap with a generator for a few weeks a year, because the last 10 percent of winter autonomy is by far the most expensive part of the system.
Does a bigger panel mean fewer panels for the same output?
For the same array kilowatts, yes, fewer modules and slightly less area, but the effect is smaller than the wattage number suggests. A 450 watt module is usually physically larger than a 400 watt one, so the watts per square foot changes much less than the watts per panel. What higher-wattage modules genuinely save is labour and racking parts: fewer mounts, fewer connectors, fewer things to go wrong. Where they cost you is flexibility on an awkward roof, where several smaller modules may fit a plane that one large one will not.
Why is my system producing less than this calculator predicted?
Check the obvious things in order. First, the peak sun hours figure you used — if it came from a national map rather than your coordinates and orientation, it is probably wrong. Second, shading, which is non-linear: a chimney shadow crossing one module for two hours can cost far more than two hours of that module. Third, temperature, if you are comparing a summer month against an annual average. Fourth, soiling, which creeps up slowly enough that nobody notices. And fifth, actual weather, because a year is not an average and a cloudy season is a real thing. Judge a system against a full twelve months, not a fortnight.