Solar Panel Size Calculator: How much solar do I need for my RV?
For 800 Wh of daily consumption at 4.5 peak sun hours (a rough annual average across the continental US), a flat-mounted panel needs around 300 W – at 3 hours in a cloudy winter month it's over 450 W, at 6 hours in the desert Southwest around 230 W. This calculator derives the right solar wattage from consumption, travel season and location so the system suffices in your worst travel month.
Result
Recommended solar power
350 W
calculated 305 W for custom value
- Calculated requirement
- 305 W
- Output of the recommendation
- 1,103 Wh/day
- Peak sun hours (reference)
- 4.5 h/day
- The system must match the battery: a sunny day with surplus only helps if the battery can absorb the energy.
Show calculation
- 1Irradiation (custom value): 4.5 peak sun hours/day
- 2Requirement: 800 Wh × (1 + 20%) ÷ (4.5 h × 70%) = 305 W
- 3Recommendation: 350 W → output 1,103 Wh/day
How it's calculated
The calculator inverts the output formula: from the desired daily output (consumption plus reserve) and the peak sun hours of the design month follows the required rated power.
Power [W] = consumption [Wh/day] × (1 + reserve) ÷ (peak sun hours × tilt factor × performance ratio)The design month decides: if you only travel in summer you get by with much less than spring and fall campers. Designing for winter is usually uneconomical in northern latitudes – then a DC-DC charger or shore power makes more sense.
- Performance ratio 60–75% for flat roof mounting (temperature, dirt, controller, cables, partial shading).
- Reserve 20% for cloudy days; if you want to bridge several rainy days you need battery capacity rather than panel area.
- The result is rounded up to common panel sizes.
Worked example
A couple with 650 Wh daily consumption travels from May to September in southern Germany; the system should suffice in May. Flat mounting, 70% performance ratio, 20% reserve.
Inputs
- Daily power consumption: 650 Wh
- Mounting: Flat on the roof
- Location: Southern Germany
- Design month: May
- Performance ratio: 70 %
- Reserve for bad days: 20 %
Result
250 W
Recommended solar power
- Calculated requirement
- 210 W
- Output of the recommendation
- 927 Wh/day
- Peak sun hours (reference)
- 5.3 h/day
Calculation
- Irradiation (May): 5.3 peak sun hours/day
- Requirement: 650 Wh × (1 + 20%) ÷ (5.3 h × 70%) = 210 W
- Recommendation: 250 W → output 927 Wh/day
The variables explained
- Daily power consumption (Wh)
- All loads including inverter losses – see the power consumption calculator.
- Design month
- The month with the lowest irradiation in which the system should still suffice.
- Peak sun hours
- Daily irradiation in kWh/m²; regional monthly values are stored for Europe, custom values are possible (e.g. from NREL PVWatts or PVGIS).
- Performance ratio (%)
- Ratio of real to theoretical output.
- Reserve (%)
- Margin for cloud cover and aging of the panels.
Common mistakes
- Designing for summer and sitting in the dark in October.
- Without battery planning: the solar power must match the battery (solar charge time and off-grid days calculators).
- Overestimating roof area: roof vents, air conditioner and satellite dish take space and cast shadows.
- PWM controllers on large systems: from about 200 W an MPPT controller almost always pays off.
Assumptions and limits
- Regional irradiation values are rounded long-term averages for Europe; the real weather deviates. For the US enter the peak sun hours for your location.
- The battery can absorb the output (sufficient capacity, not full, temperature within the charging range).