Solar Panel Tilt Angle Calculator

Solar Panel Tilt Angle Calculator

The best fixed tilt for your latitude, and — computed rather than asserted — the share of a year’s irradiation your actual tilt and compass direction capture against it. With the summer and winter angles, a monthly yield curve, and the cost of facing the wrong way worked out separately from the cost of the wrong tilt.

Fixed tilt and what it captures

Latitude, tilt and aim → share of the best year
Positive north of the equator, negative south of it. The page works out which way the array should face from the sign.
0 is flat, 90 is vertical. A 4:12 roof pitch is 18.4°, 5:12 is 22.6°, 6:12 is 26.6°.
0 means due south in the northern hemisphere, due north in the southern. ±90 is due west or due east; ±180 faces away from the equator.
0.30 is typical for a sunny mid-latitude site, 0.40–0.55 for a cloudy maritime one, 0.20–0.25 for a desert. Take it from your own site data — this page does not ship an irradiance table.
0.2 for grass, soil or asphalt; 0.5–0.8 for fresh snow; about 0.1 for dark water.
A geometry, not a schematic: a section through the array, looking along the equator-facing direction. The ground is horizontal, the module is the line tilted up away from the equator side, and the two arrows are the noon sun at the summer and winter solstices, 23.4 degrees either side of its equinox height – which is exactly why one fixed tilt has to be a compromise. The plan view on the right is the azimuth: the arrow is the direction the module faces, measured from the equator-facing direction. Nothing moves, because a geometry has no loop.
96.09%Example

40° N, a 22° roof facing 30° west of south, diffuse fraction 0.30, ground reflectance 0.20

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Irradiation on a tilted plane, integrated over the year

cos θ = a + b·cos ω + c·sin ω,   with a = sin δ (sin φ cos β − cos φ sin β cos γ),   b = cos δ (cos φ cos β + sin φ sin β cos γ),   c = cos δ sin β sin γ
the surface is lit where |ω − ω0| < arccos(−a ÷ √(b²+c²)), ω0 = atan2(c, b); the day’s beam is ∫(a + b cos ω + c sin ω) dω over that window ∩ [−ωs, ωs]
plane of array = (1 − fd)·beam + [fd(1 + cos β)/2 + ρ(1 − cos β)/2]·horizontal
φ, β, γ
latitude, tilt from horizontal, and the surface’s azimuth from the equator-facing direction
δ
the sun’s declination on the day: 23.45° sin(360(284 + n) ÷ 365)
ω s
the sunset hour angle, arccos(−tan φ tan δ). 15° is one hour
f d
the diffuse share of the global horizontal irradiation. The (1 ± cos β)/2 terms are the sky and ground view factors of the tilted plane

Worked example

40° N, a 22° roof facing 30° west of south, diffuse fraction 0.30, ground reflectance 0.20
The model integrates the beam on the plane over each of twelve representative days (17, 47, 75, 105, 135, 162, 198, 228, 258, 288, 318, 344) and adds the isotropic sky and ground-reflected terms
Searching equator-facing tilts, the best fixed angle here is 36.3° — 0.91 times the latitude, not equal to it, because the diffuse component favours a flatter plane
Your 22° roof, if it faced due south, would capture 97.68% of that optimum: the tilt alone costs 2.32 points
Turning it 30° to the west costs a further 1.59 points, giving 96.09%
For comparison, the same modules lying flat would capture 85.51%, so the roof pitch is worth 10.58 points over flat
Multiply your site's own measured annual irradiation by 0.9609 to turn this into kilowatt-hours

Tilt, facing the equator, at 40° N

TiltAnnual captureShortfall
0°85.51%14.49%
10°92.24%7.76%
20°96.98%3.02%
22°97.68%2.32%
30°99.55%0.45%
36°100.00%-0.00%
45°99.13%0.87%
60°93.64%6.36%
90°69.57%30.43%
Computed by this page’s own model. The curve is remarkably flat near the top: anything from about 25° to 47° is within 1% of the best, which is why a roof pitch almost never needs changing. Flat and vertical are the expensive mistakes.

Facing the wrong way, at the optimum tilt, 40° N

Degrees off the equator-facing directionAnnual captureShortfall
0°100.00%-0.00%
15°99.50%0.50%
30°98.01%1.99%
45°95.50%4.50%
60°91.95%8.05%
90°81.93%18.07%
135°62.57%37.43%
180°52.47%47.53%
Also flat near the top, and for the same reason: a surface tilted 36° still sees most of the sky whichever way it points. Thirty degrees off costs about 2%; due east or due west costs about 15%; facing away from the equator at a steep tilt is what really hurts.

Optimum tilt against latitude, and the seasonal angles

LatitudeAnnual optimumAs a fraction of latitudeSummer tiltWinter tilt
0.0°0.5°—0.0°15.5°
10.0°9.4°0.940.0°24.4°
20.0°18.3°0.923.3°33.3°
30.0°27.3°0.9112.3°42.3°
40.0°36.3°0.9121.3°51.3°
50.0°45.1°0.9030.1°60.1°
60.0°53.2°0.8938.2°68.2°
-23.5°20.4°0.875.4°35.4°
-33.9°29.7°0.8814.7°44.7°
-45.0°39.7°0.8824.7°54.7°
The classic rule is “tilt equals latitude”, and it is exactly right for a geometric beam-only model — but adding a real diffuse fraction flattens the optimum to roughly 0.9 of the latitude at mid-latitudes. The summer and winter columns are the usual ∓15° rule of thumb applied to the computed optimum, not separate optimisations.

The example roof, month by month

MonthRepresentative dayDeclinationHorizontalYour arrayDifference
Jan17-20.92°53.879.225.4
Feb47-12.95°72.795.222.5
Mar75-2.42°97.1113.616.5
Apr1059.41°122.5129.97.3
May13518.79°140.4139.0-1.4
Jun16223.09°147.7141.9-5.8
Jul19821.18°143.9140.0-3.8
Aug22813.45°129.2132.93.7
Sep2582.22°106.1119.113.0
Oct288-9.60°79.7100.320.6
Nov318-18.91°58.282.824.7
Dec344-23.05°48.774.525.8
Both columns are indexed so that the average horizontal month is 100. Tilting up flattens the seasonal swing: the horizontal surface runs from 49 in December to 148 in June, the 22° roof from 75 to 142. That flattening, not the annual total, is usually what a tilt is for.

Why tilt equals latitude is nearly right, and what nearly costs

A fixed array is a compromise between a summer sun that climbs high and a winter sun that stays low. The usual advice is to tilt it at the latitude, and there is a clean reason for that: for a surface facing the equator, tilting by β makes the geometry identical to a horizontal surface at latitude φ − β, and a horizontal surface collects the most beam over a year at the equator. So for pure beam radiation the annual optimum is exactly the latitude. PVEducation states the rule in those words, and this page’s model reproduces it when the diffuse fraction is set to zero.

Diffuse light flattens it. Real skies are not beam-only. A tilted plane sees only (1 + cos β)/2 of the sky dome and picks up ρ(1 − cos β)/2 from the ground, so tilting always throws some diffuse light away. With a 30% diffuse fraction and grass underneath, the optimum at 40° comes out at about 36°, roughly 0.9 of the latitude — which is where the various “0.87 × latitude” rules of thumb come from. A cloudy site with a 50% diffuse fraction optimises lower still; a desert site higher.

The curve near the top is almost flat. That is the practical result, and it is the reason to compute the loss instead of arguing about the angle. At 40° N, anything between about 25° and 47° captures within 1% of the best. Thirty degrees of azimuth error costs about 2%. It takes a flat roof, a vertical wall or a genuinely east- or west-facing pitch to lose double digits. A roof that is structurally 22° and points 30° west is at 96% of the ideal, and no tilt frame will ever pay for the missing 4%.

Seasonal adjustment. If the array is on a ground frame and you are willing to move it, the usual schedule is the annual optimum minus about 15° for summer and plus about 15° for winter, with the annual angle for spring and autumn. A two-season change captures most of what a four-season change does. Whether it is worth doing depends entirely on the load: for a summer-heavy load a flatter fixed angle is better than any schedule, and for an off-grid system sized by the WORST month, a steep winter-biased fixed tilt beats the annual optimum even though it collects less over the year. The monthly chart on this page is the one to read for that, not the annual percentage.

What this model does and does not do. It computes a RATIO between two orientations at one site, from geometry plus an isotropic sky. It does not know your weather. To get kilowatt-hours, take your site’s measured annual global horizontal irradiation — from a national meteorological service or an atlas such as the Global Solar Atlas — and multiply. It assumes a clear horizon and no shading, which is often the largest real effect and is entirely absent here; a single chimney can cost more than every angle on this page. It uses an isotropic sky, which understates a surface aimed near the sun by a few per cent against an anisotropic model such as Perez’s. It ignores snow, soiling, the incidence-angle reflection loss at the glass, and the module’s own temperature coefficient. And it samples the year at twelve days rather than 365: checked against a full 365-day integration, the capture percentage agrees within about 0.3 points at the worst combination tried.

For the string and inverter side of the same array, the solar string sizing calculator; for sizing the array against a daily load, the solar panel size calculator; and for the controller, the solar charge controller calculator.

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Frequently asked questions

What angle should solar panels be at?

Close to the latitude, and a little flatter. At 40° the best fixed angle with a typical 30% diffuse fraction is about 36°. The curve is flat near the top, so anything within about 10° of that loses less than 1%.

How much do I lose if my roof faces south-west instead of south?

Very little. At 40° N and the optimum tilt, 30° off the equator-facing direction costs about 2% of the annual irradiation, and 45° costs about 4%. Due east or due west costs around 15%.

Is tilt equal to latitude actually correct?

It is exactly correct for a beam-only geometric model, because tilting a south-facing surface by β makes it behave like a horizontal surface at latitude φ − β. Adding diffuse light, which a tilted plane sees less of, moves the real optimum down to roughly 0.9 of the latitude.

How much does a flat roof lose?

At 40° N with a 30% diffuse fraction, about 14% of what a well-aimed fixed array captures. Nearer the equator it is much less — at 10° latitude, under 2% — and further towards the poles much more.

Should I adjust the tilt seasonally?

It gains a few per cent a year: the usual schedule is the annual optimum minus 15° for summer and plus 15° for winter. Whether that is worth the trips outside depends on the array. It matters far more for an off-grid system sized by its worst month, where a winter-biased fixed tilt can be the right answer even though it collects less over the year.

Why does this page ask for a diffuse fraction?

Because it is the input that moves the optimum tilt. A tilted surface sees less of the sky dome, so the more of your light arrives as diffuse skylight rather than direct beam, the flatter the best tilt becomes. Use your own site’s figure; this page deliberately ships no irradiance table.

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References

  1. Honsberg C, Bowden S. PVCDROM — Solar Radiation on a Tilted Surface and Declination Angle, pveducation.org. The declination equation used here is theirs, δ = 23.45° sin[(360 ÷ 365)(284 + d)], and they state the rule this page tests: “for a fixed tilt angle, the maximum power over the course of a year is obtained when the tilt angle is equal to the latitude of the location”.
  2. Duffie JA, Beckman WA. Solar Engineering of Thermal Processes, 4th ed. Wiley, 2013. §2.15 “Radiation on Sloped Surfaces: Isotropic Sky” and its equation 2.15.1 — the (1 + cos β)/2 sky and (1 − cos β)/2 ground view factors used here. Section numbers confirmed against the publisher’s own chapter contents; the chapter-1 table of monthly average days was NOT fetched, so this page lists the twelve day numbers it uses (17, 47, 75, 105, 135, 162, 198, 228, 258, 288, 318, 344) in the worked example, where they can be checked directly.
  3. Liu BYH, Jordan RC. The long-term average performance of flat-plate solar-energy collectors. Solar Energy 1963;7(2):53–74. The isotropic sky model. Volume, year and page range confirmed; the paper itself was not fetched, and what was checked instead is numerical: the closed-form beam integral here agrees with a 6,000-step numerical integration of cos θ over the lit window to better than 1e-5 across 4,000 random latitude, tilt, azimuth and day combinations.
  4. World Bank / ESMAP / Solargis. Global Solar Atlas (globalsolaratlas.info) for the site’s own measured global horizontal irradiation and diffuse fraction, which this page asks you to supply rather than shipping a copyrighted table.