Sun Path Geometry for Solar: Tilt, Azimuth and Row Spacing from First Principles

A flat-roof layout drawing carries numbers that look arbitrary: a tilt of 10°, a row pitch of 2.7 m, a 1.2 m setback from the parapet. Each follows from where the sun is on a given day and hour, and that can be calculated from latitude, date and time with a few trigonometric formulas.

This guide builds those formulas from first principles. We fix the conventions first, derive the sun's altitude and azimuth from one vector dot product, and from it get day length, the angle at which light meets a tilted module, the effect of tilt on annual beam energy and the spacing between rows. The worked examples use Tirupati (about 13.63° N, 79.42° E) and every figure was computed in Python. For the energy side, continue to our guide to solar energy yield and performance ratio.

Angles, symbols and sign conventions

Solar geometry usually goes wrong through sign conventions, not trigonometry. They are fixed here, and every symbol is defined before use.

• $\varphi$ is the latitude in degrees, positive north. For Tirupati, $\varphi = 13.63^{\circ}$.
• $L_{\text{loc}}$ is the site longitude in degrees east (79.42° for Tirupati). $L_{\text{st}}$ is the reference meridian of the time zone: 82.5° E for Indian Standard Time (IST).
• $n$ is the day number of the year, with 1 January as 1.
• $\delta$ is the solar declination in degrees: the angle between the sun's rays and the equatorial plane, positive when the sun is north of the equator.
• $\omega$ is the hour angle in degrees: zero at solar noon, negative before it, positive after it, changing 15° every hour.
• $\alpha$ is the solar altitude above the horizontal. The zenith angle is $\theta_z = 90^{\circ} - \alpha$.
• $\gamma_s$ is the solar azimuth from due south, east negative and west positive. A south-east sun has $\gamma_s$ near −45°. Add 180° for a compass bearing.
• $\beta$ is the module tilt from horizontal and $\gamma$ the azimuth it faces, with the same signs as $\gamma_s$. A module facing due south has $\gamma = 0^{\circ}$.
• $\theta$ is the angle of incidence between a sun ray and the perpendicular (normal) to the module.

We assume a point sun with parallel rays, no atmospheric bending (under 0.6° even at the horizon) and a flat, level roof. The resulting angles are accurate to a fraction of a degree.

Why the sun's path changes: declination

The Earth's axis is tilted about 23.44° from the perpendicular to its orbit and points in a nearly fixed direction in space, so over a year the sun moves north and south of the equatorial plane. On about 20 March and 23 September $\delta = 0^{\circ}$; on 21 June $\delta = +23.44^{\circ}$; on 21 December $\delta = -23.44^{\circ}$. Cooper's approximation (1969) gives it from the day number alone:

The sine captures the smooth annual swing and the 284-day offset puts the zero crossing at the March equinox. Against the NOAA astronomical series, the largest difference over the year is 0.98°, in mid-October, so treat it as good to about a degree.

Worked example. For 16 October, $n = 289$. The angle inside the sine is $360^{\circ}\times 573/365 = 565.15^{\circ}$, which is $205.15^{\circ}$ after removing one full turn. Then $\sin 205.15^{\circ} = -0.4250$, so:

The sun is about 10° south of the equator, as expected a few weeks after the September equinox. NOAA gives −8.99°, which is the one-degree error mentioned. Tirupati's latitude is 13.63° N, so the sun is directly overhead at noon twice a year, when $\delta = \varphi$: around 27 April and 15 August. Between those dates the noon sun is north of you, which matters for water-tank shadows later.

Solar time versus clock time

Solar time is set by the sun, with noon when it crosses your meridian. Your clock shows standard time for a whole country. Define $t_{\text{clock}}$ and $t_{\text{solar}}$ as clock and solar time in minutes after midnight, and $E$ as the equation of time in minutes, the difference between apparent and uniform mean time. Then:

The factor 4 arises because the Earth turns 15° per hour, which is 4 minutes per degree. Tirupati is 3.08° west of the 82.5° E meridian, so the longitude term is $4\times(79.42-82.5) = -12.32$ minutes: solar noon comes 12.32 minutes after clock noon before the second correction.

The equation of time arises because the orbit is an ellipse and the axis is tilted. Spencer's approximation (1971) uses an angle $B$ that advances about 1° a day:

The result is in minutes (229.18 is 1440 minutes divided by $2\pi$). Against NOAA it stays within about 0.8 minutes all year.

Setting $t_{\text{solar}} = 12{:}00$ gives the clock time of solar noon:

Worked example, 16 October. With $n = 289$, $B = 284.05^{\circ}$ and $E = +14.62$ minutes. Then $t_{\text{noon}} = 12{:}00 + 12.32 - 14.62$ min, which is 11:57:42 IST. A clock reading of 10:00 IST is a solar time of $600 - 12.32 + 14.62 = 602.30$ minutes, or 10:02:18, so the hour angle is

The negative sign confirms that this is before solar noon. The table repeats the calculation for other dates.

A shading study quoted in IST hours must apply both corrections, since 30 minutes of error moves the sun by 7.5°. Our Tirupati solar guide covers the local climate and approvals.

Solar altitude and azimuth from vectors

We want the altitude $\alpha$ from latitude, declination and hour angle. The cleanest route uses unit vectors (arrows of length 1). Use an Earth-fixed frame: $z$ along the rotation axis towards the north pole, $x$ from the Earth's centre to where the observer's meridian crosses the equator, and $y$ towards the east. The local vertical makes angle $\varphi$ with the equatorial plane and has no east component. The sun makes angle $\delta$ with that plane and lies $\omega$ west of the meridian, so in the morning (negative $\omega$) it has a positive east component. That gives:

Here $\hat{\mathbf{u}}$ points up and $\hat{\mathbf{s}}$ points to the sun. The dot product of two unit vectors is the cosine of the angle between them, and that angle is the zenith angle, with $\cos\theta_z = \sin\alpha$. So:

At solar noon ($\omega = 0$) this reduces to $\alpha = 90^{\circ} - \varphi + \delta$. At Tirupati on 21 December the noon altitude is $90 - 13.63 - 23.45 = 52.92^{\circ}$; at the equinoxes it is $76.37^{\circ}$.

For the azimuth we need the sun's south and west components. Dotting $\hat{\mathbf{s}}$ with the local east vector $(0,1,0)$ and north vector $(-\sin\varphi, 0, \cos\varphi)$, and reversing the sign of the east and north results, gives:

These equal $\cos\alpha$ times the cosine and sine of the azimuth. Dividing by $\cos\delta$ gives the working form, where atan2 is the two-argument arctangent that returns the correct quadrant:

Atan2 matters here: between late April and mid-August the noon sun at Tirupati is north of you, and an arcsine formula cannot tell a south-east sun from a north-east one.

Worked example. Take 16 October at 10:00 IST: $\omega = -29.42^{\circ}$, $\delta = -9.97^{\circ}$, $\varphi = 13.63^{\circ}$. Then $\sin\varphi = 0.2357$, $\cos\varphi = 0.9718$, $\cos\delta = 0.9849$, $\cos\omega = 0.8710$, and

so $\alpha = 52.46^{\circ}$ and $\theta_z = 37.54^{\circ}$. For the azimuth, $\sin\omega = -0.4913$ and $\sin\varphi\cos\omega - \cos\varphi\tan\delta = 0.2053 + 0.1708 \approx 0.376$, so $\gamma_s = -52.57^{\circ}$: a compass bearing of 127.4°, south-east. A south-east sun about 52° up at 10 in the morning in mid-October is what you would expect.

Sunrise, sunset and day length

At sunrise and sunset $\alpha = 0$. Writing $\omega_s$ for the sunset hour angle in degrees, the altitude equation gives

The sun is up from $-\omega_s$ to $+\omega_s$, and at 15° per hour the day length $T_d$ in hours is

This holds when $|\tan\varphi\tan\delta| \lt 1$, always true in India. For Tirupati $\tan\varphi = 0.2425$. On 21 December $\tan\delta = -0.4338$, so $\cos\omega_s = 0.1052$, $\omega_s = 83.96^{\circ}$ and $T_d = 2\times 83.96/15 = 11.19$ h (11 h 12 min). The table uses the idealised $\delta = 0^{\circ}$ and $\pm 23.45^{\circ}$. The refraction column allows for atmospheric bending and the sun's disc, which add about 7 to 8 minutes.

Sanity check: the equinox day is 12 hours at any latitude. Summer and winter differ by only 1.6 hours at Tirupati, so the season changes the sun's height far more than the day length.

Tilted surfaces and the angle of incidence

A beam of 1 m² cross-section hitting a module at angle $\theta$ from the normal spreads over $1/\cos\theta$ square metres, so each square metre receives $\cos\theta$ times the intensity. With $G_{bn}$ the beam irradiance on a surface square to the sun, in W/m², the beam power per unit module area is

This is the cosine loss: 86.6% of the full beam at $\theta = 30^{\circ}$, 50% at $60^{\circ}$. In local south, west and up coordinates the sun is $(\cos\alpha\cos\gamma_s,\ \cos\alpha\sin\gamma_s,\ \sin\alpha)$ and the module normal is $(\sin\beta\cos\gamma,\ \sin\beta\sin\gamma,\ \cos\beta)$. Their dot product gives the general formula:

It holds while the sun is above the horizon and $\cos\theta \gt 0$; a negative value means the sun is behind the module and the beam contribution is zero.

The equivalent-latitude shortcut for south-facing modules

For a south-facing module there is a neater route. Tilting it by $\beta$ turns its normal from the local vertical $(\cos\varphi, 0, \sin\varphi)$ towards the equator, giving $(\cos(\varphi-\beta), 0, \sin(\varphi-\beta))$: the vertical of a site at latitude $\varphi-\beta$. Dotting with $\hat{\mathbf{s}}$:

This is the altitude formula with $\varphi$ replaced by $\varphi-\beta$. A tilt equal to the latitude makes the equivalent latitude zero, so $\cos\theta = \cos\delta\cos\omega$ and the noon angle of incidence is $|\delta|$, never above 23.45°.

Worked example. Same instant as before (16 October, 10:02 solar time), with $\beta = 13.6^{\circ}$ so $\varphi - \beta = 0.03^{\circ}$. Then $\cos\theta \approx \cos\delta\cos\omega = 0.9849\times 0.8710 = 0.858$, $\theta = 30.9^{\circ}$. The flat surface has $\cos\theta_z = 0.7929$, so the tilted module collects $0.858/0.793 = 1.082$ times as much beam per square metre, about 8% more. The general formula with $\alpha = 52.46^{\circ}$ and $\gamma_s = -52.57^{\circ}$ returns the same 0.858.

At noon the seasonal effect is stronger. On 21 December a flat surface has $\cos 37.08^{\circ} = 0.798$ and the 13.6° module $\cos 23.48^{\circ} = 0.917$, a gain of 15.0%. On 21 June the sun is north of the zenith: flat gives $\cos 9.82^{\circ} = 0.985$ and tilted $\cos 23.42^{\circ} = 0.918$, a loss of 6.9%. A fixed tilt trades summer beam for winter beam.

Choosing the tilt angle

To compare tilts we sum the beam power on the plane over every daylight minute of the year. The model is deliberately simple:

• Every day is clear and only beam light counts; diffuse light, clouds, temperature and soiling are ignored.
• Beam irradiance square to the sun is $G_{bn} = 1.353\times 0.7^{\,AM^{0.678}}$ kW/m², a standard clear-sky relation. Here $AM = 1/\sin\alpha$ is the air mass, 1.353 an older value of the solar constant and 0.7 the transmittance per air mass.
• Declination comes from Cooper's formula for each day $n$, with one-minute steps and the sun counted above 1° altitude.

$H$ is the annual beam energy per square metre of south-facing plane, in kWh/m². The absolute values are a clear-sky ceiling, so only the ratios between tilts carry meaning. The last column, $(1+\cos\beta)/2$, is the share of isotropic sky-diffuse light a tilted plane keeps relative to a flat one.

The curve is very flat. The model's maximum lies near 12°, about 1.6° below the latitude, and every tilt from 8° to 16° is within 0.3% of it. Moving from flat to 13.6° adds 2.2% a year, raises the 21 December total by 19% and lowers the 21 June total by about 11%. A 30° tilt gains 34% on 21 December, gives up 30% on 21 June and ends 4.8% behind on the year.

Why 10° to 15° is the usual rooftop choice

Because yield barely changes, other factors decide. A tilt of 8° to 10° or more lets rain wash dust off the glass, while lower tilts generally present less area to the wind and allow closer rows. Our Tirupati guide suggests 13° to 15° as a starting layout and our system sizing guide 12° to 15°.

Shadows and row spacing

On a flat roof each row must avoid shading by the row in front during the hours that produce most of the energy. Define $l$ as the module slant length in metres (the long side in portrait) and $\beta$ as the tilt. The top edge of a row then stands $h = l\sin\beta$ above its lower edge, and the row has horizontal depth $l\cos\beta$. The row pitch $P$ is the north-south distance between the lower edges of successive rows.

The length of a shadow

A point at height $h$ casts its shadow along the sun's ray, which descends at angle $\alpha$. By similar triangles the shadow on level ground has length $L = h/\tan\alpha$ and points away from the sun. Resolving it along north and east with the solar azimuth:

$x_N$ is how far north the shadow tip lands and $x_E$ its offset to the east, negative (west) when the sun is in the east. In vector form $x_N = h\,s_S/\sin\alpha$ and $x_E = h\,s_W/\sin\alpha$. A south-east or south-west sun spreads the length sideways, so the north reach is shorter than $L$.

Pitch for no shading from 9:00 to 15:00

The top edge's shadow lands $x_N$ north of it, and the next row's lower edge is $P - l\cos\beta$ north of that edge. For no shading the gap must be at least $x_N$ at the worst moment:

The reach is longest at the window edges, on the day the noon sun is lowest: the winter solstice. Using 9:00 to 15:00 solar time on 21 December is a common design rule, but it is a design choice, not a law of nature. In the beam model above, about 78% of the 21 December beam energy on a 10° plane arrives inside that window; the first and last hour may see some shading.

The ground coverage ratio, the fraction of the roof strip occupied by module area, is

Worked example. Modules 2.28 m long in portrait at $\beta = 10^{\circ}$, at Tirupati on 21 December. At 9:00, $\omega = -45^{\circ}$ and $\delta = -23.45^{\circ}$. The altitude equation gives $\sin\alpha = -0.0938 + 0.6304 = 0.5367$, so $\alpha = 32.46^{\circ}$. The azimuth is $\gamma_s = -50.25^{\circ}$, so $\cos\gamma_s = 0.6395$. The height of the top edge is

The shadow length is $L = 0.396/\tan 32.46^{\circ} = 0.396/0.6360 = 0.6225$ m, and its north reach is $x_N = 0.6225\times 0.6395 = 0.398$ m. The depth of the row is $2.28\cos 10^{\circ} = 2.245$ m. Then:

At 12:00 the reach is only 0.299 m, so the window edges do set the pitch. Errors in $\delta$ matter little: Cooper's value is accurate to 0.01° on 21 December, and even a full degree would change this pitch by under 2 cm. The table repeats the calculation for other tilts (2.28 m modules, 9:00 to 15:00).

Each extra 5° of tilt costs about 0.13 to 0.17 m of pitch. The window matters too. At 10° tilt, 10:00 to 14:00 needs a 2.58 m pitch (GCR 0.884), 9:00 to 15:00 needs 2.64 m (0.863) and 8:00 to 16:00 needs 2.80 m (0.814). Yield is flat across tilts but pitch is not, so a lower tilt packs more modules into a roof.

For scale, an illustrative 1.13 m wide, 550 W module at this pitch takes 3.00 m² of roof strip, or 5.4 m² per kWp (about 59 sq ft). That is the bare array field; the 80 to 100 sq ft per kW rule in our sizing guide also covers walkways and setbacks. Use your own datasheet.

Parapets and water tanks on a terrace

The same shadow formulas apply to any obstruction. For one of height $H$ above the roof, with the lower edge of the nearest module raised $z_0$ above the roof, the module is shade-free at distance $d$ north of it if

This is the pitch equation with $H - z_0$ in place of $h$. It is conservative, since the module surface rises away from its lower edge, the hardest point to clear.

A 1 m parapet

Take a 1 m parapet on the south edge and a module lower edge at roof level ($z_0 = 0$). On 21 December the formula gives 1.005 m at 9:00 and 15:00, 0.846 m at 10:00 and 14:00, 0.776 m at 11:00 and 13:00 and 0.756 m at noon. The first row therefore starts about 1.0 m north of the parapet. Raising the lower edge 0.3 m cuts the setback to about 0.70 m. East and west parapets behave differently: at 9:00 the shadow of a 1 m wall reaches 1.209 m westward ($x_E$), so keep about 1.2 m clear of the east parapet in the morning and of the west parapet at 15:00.

A 1.8 m water tank

A 1.8 m tank is more serious. On 21 December its shadow tip lies 1.81 m north and 2.18 m west at 9:00, 1.36 m north at noon, and 1.81 m north and 2.18 m east at 15:00. Anything in that fan, plus the tank's width, is shaded at some hour. Around 21 June the noon sun is north of the zenith and the shadow falls south, only 0.31 m long at noon and 0.58 m at 9:00. So put the tank on the north side of the array: its winter shadow then points away from the modules and its summer shadow is short.

Electrically, a shadow on one module can limit the current of its whole string, so a shade survey should give measured heights, a plotted shadow envelope for the design window and strings assigned so that shaded modules sit together. Take measurements from safe positions away from unprotected terrace edges, and remember that PV strings are live whenever light falls on them.

Using this on your own roof

The workflow is short. Fix a tilt of 10° to 15°, compute the winter-solstice altitude and azimuth at the edges of your window (32.46° and 50.25° at 9:00 and 15:00 in Tirupati), add the shadow reach of the top edge to the row depth for the pitch, and check each obstruction with the setback formula. A ten-line script reproduces every table here.

These are first-order results; real designs add walkways, wind and structural limits and a loss simulation. If you would like this done for your own terrace, with the shadow envelope of your tank and parapets drawn on a layout, book a free power audit with Ray2Volt.

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