Latitude Radiation Budget Calculator
Our planetary & earth system science calculator computes latitude radiation budget accurately. Enter measurements for results with formulas and error
Reviewed for accuracy by Daniel Agrici, Founder & Lead Developer
Latitude Radiation Budget Calculator
Calculator
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Formula: x = sin(lat); P2 = (3x^2 - 1)/2; Q = (S0/4)(1 - S2*P2); ASR = A0(1 - B2*P2); OLR = L0 + L2*P2; R_net = ASR - OLR; F = pi*R_E^2*(A0*B2 + L2)*โฆ
Worked example โ Q: 414.06 W/m2 | ASR: 317.33 W/m2 | OLR: 267.29 W/m2 | Net: +50.04 W/m2 surplus | Transport: 2.37 PW northward
Formula
x = sin(lat); P2 = (3x^2 - 1)/2; Q = (S0/4)(1 - S2*P2); ASR = A0(1 - B2*P2); OLR = L0 + L2*P2; R_net = ASR - OLR; F = pi*R_E^2*(A0*B2 + L2)*x(1 - x^2)
Two-mode Legendre energy balance model in x = sin(latitude), with P2(x) = (3x^2 - 1)/2. S0 is the solar constant (1361 W/m2) and Qbar = S0/4 = 340.25 W/m2 is its global mean. S2 = 0.477 is the annual mean insolation profile factor obtained by integrating the daily insolation formula over a year at obliquity 23.44 degrees and projecting onto P2. A0 = (S0/4)(1 - albedo) = 239.88 W/m2 is global mean absorbed shortwave, and B2 = 0.71 is the absorbed-shortwave profile factor; B2 exceeds S2 because planetary albedo also rises poleward, steepening the absorbed flux by about 50 percent relative to the insolation gradient alone. Outgoing longwave is written as OLR = L0 + L2*P2 with L0 = 240 W/m2 and L2 = -60 W/m2. Local planetary albedo is recovered as 1 - ASR/Q. Poleward transport comes from integrating the net flux anomaly over the spherical cap, F(phi) = 2 pi R_E^2 times the integral of (R_net - global mean) dx from -1 to sin(phi); with R_E = 6.371e6 m this reduces exactly to F = pi R_E^2 (A0*B2 + L2) x(1 - x^2), peaking at x = 1/sqrt(3), i.e. 35.26 degrees. The model reproduces observed CERES fluxes to roughly 10 percent; it cannot capture the subtropical maximum or equatorial dip in outgoing longwave, which need higher Legendre modes.
Worked Examples
Example 1: Tropical Surplus at 10 Degrees North
Problem:Find the annual mean radiation budget and implied poleward transport at 10 degrees N using S0 = 1361 W/m2, S2 = 0.477, global albedo 0.295, B2 = 0.71, L0 = 240 W/m2, L2 = -60 W/m2.
Solution:x = sin(10 deg) = 0.173648, x^2 = 0.030154 P2 = (3(0.030154) - 1)/2 = -0.454769 Qbar = 1361/4 = 340.25 W/m2, A0 = 340.25(1 - 0.295) = 239.876 W/m2 Q = 340.25[1 - 0.477(-0.454769)] = 340.25(1.216925) = 414.06 W/m2 ASR = 239.876[1 - 0.71(-0.454769)] = 239.876(1.322886) = 317.33 W/m2 Implied albedo = 1 - 317.33/414.06 = 0.234 OLR = 240 + (-60)(-0.454769) = 240 + 27.29 = 267.29 W/m2 R_net = 317.33 - 267.29 = +50.04 W/m2 (surplus) A0*B2 + L2 = 239.876(0.71) - 60 = 170.312 - 60 = 110.312 W/m2 x(1 - x^2) = 0.173648(0.969846) = 0.168412 pi R_E^2 = pi (6.371e6 m)^2 = 1.27516e14 m^2 F = 1.27516e14 * 110.312 * 0.168412 = 2.369e15 W
Result:Q: 414.06 W/m2 | ASR: 317.33 W/m2 | OLR: 267.29 W/m2 | Net: +50.04 W/m2 surplus | Transport: 2.37 PW northward
Example 2: Polar Deficit at 70 Degrees North
Problem:Repeat the calculation at 70 degrees N with the same seven inputs to see the high-latitude deficit.
Solution:x = sin(70 deg) = 0.939693, x^2 = 0.883022 P2 = (3(0.883022) - 1)/2 = +0.824533 Q = 340.25[1 - 0.477(0.824533)] = 340.25(0.606698) = 206.43 W/m2 ASR = 239.876[1 - 0.71(0.824533)] = 239.876(0.414581) = 99.45 W/m2 Implied albedo = 1 - 99.45/206.43 = 0.518 OLR = 240 + (-60)(0.824533) = 240 - 49.47 = 190.53 W/m2 R_net = 99.45 - 190.53 = -91.08 W/m2 (deficit) x(1 - x^2) = 0.939693(0.116978) = 0.109923 F = 1.27516e14 * 110.312 * 0.109923 = 1.546e15 W
Result:Q: 206.43 W/m2 | ASR: 99.45 W/m2 | OLR: 190.53 W/m2 | Net: -91.08 W/m2 deficit | Transport: 1.55 PW northward
Example 3: Peak Poleward Energy Transport
Problem:Find the maximum poleward energy transport and the latitude at which it occurs, using the same defaults.
Solution:F(phi) = pi R_E^2 (A0*B2 + L2) x(1 - x^2), so dF/dx = 0 when 1 - 3x^2 = 0 x = 1/sqrt(3) = 0.577350, latitude = arcsin(0.577350) = 35.26 deg x(1 - x^2) = 0.577350(1 - 1/3) = 2/(3 sqrt 3) = 0.384900 A0*B2 + L2 = 110.312 W/m2 (as above) F_max = 1.27516e14 m^2 * 110.312 W/m2 * 0.384900 = 5.414e15 W At that same latitude P2 = 0, so R_net = A0 - L0 = 239.876 - 240 = -0.12 W/m2, essentially the zero crossing
Result:Peak transport: 5.41 PW at 35.3 degrees | Zero crossing of net radiation: 35.2 degrees
Frequently Asked Questions
How does solar radiation vary with latitude?
Solar radiation intensity decreases from the equator toward the poles because of the cosine law of illumination: the flux received by a surface is proportional to the cosine of the solar zenith angle. At the equator, sunlight strikes nearly perpendicular to the surface, concentrating energy over a small area. At 60 degrees latitude the same beam is spread over twice the area, delivering only half the flux. At the poles during solstice the sun never rises high in the sky, and during the polar night no direct sunlight arrives at all. This latitudinal gradient is the primary driver of atmospheric and oceanic circulation.
How does the cosine law govern solar radiation at each latitude?
The cosine law states that the irradiance on a surface equals the solar constant multiplied by the cosine of the angle between the incoming beam and the surface normal, which at a horizontal surface equals the cosine of the solar zenith angle. Mathematically, E equals S times cos(theta_z), where theta_z is the zenith angle. When the sun is directly overhead theta_z is zero, cos is one, and irradiance equals the full solar constant. As the sun moves toward the horizon theta_z approaches 90 degrees, cos approaches zero, and irradiance drops to nearly nothing. This relationship governs the enormous difference in annual insolation between tropical and polar regions.
What is the Legendre profile used to model insolation by latitude?
Annual mean insolation at the top of the atmosphere is conventionally written as Q(x) = (S0/4) times [1 - S2 P2(x)], where x is the sine of latitude and P2(x) = (3x^2 - 1)/2 is the second Legendre polynomial. Integrating the exact daily insolation formula over a full year for a circular orbit at Earth's present obliquity of 23.44 degrees and projecting the result onto P2 gives S2 = 0.477, the value used as the default here. The global mean of P2 is zero, so the global mean of Q is exactly S0/4, or 340.25 W/m2 for a solar constant of 1361 W/m2. The two-term truncation gives 421 W/m2 at the equator and 178 W/m2 at the pole against exact values of 416 and 172 W/m2, so it is accurate to about 1 to 3 percent, with the residual carried by the neglected fourth-order term.
Why do polar latitudes receive less annual solar radiation than the tropics?
Polar latitudes receive less annual solar energy for two compounding reasons. First, the low sun angle spreads each beam of sunlight over a larger surface area, reducing flux by the cosine of the zenith angle. Second, polar regions experience an extended polar night during which the sun stays below the horizon entirely and insolation drops to zero, lasting close to six months at the poles themselves and shortening to a single day at the Arctic and Antarctic circles. Although polar summers have 24-hour daylight that partially compensates, the continuously low sun angle limits instantaneous flux so severely that total annual insolation remains far below equatorial values. Exact integration of the daily insolation formula gives about 416 W/m2 annually at the equator against about 172 W/m2 at the pole, a ratio near 2.4. The combination of these geometric effects creates a surplus of radiation in the tropics and a deficit at the poles that drives global heat transport.
What is the seasonal variation of insolation at different latitudes?
Seasonal insolation variation is smallest at the equator and largest at the poles. At the equator, the sun passes nearly overhead year-round and day length stays close to 12 hours, so insolation fluctuates only modestly between solstices, with a weak semiannual double peak at the equinoxes. At mid-latitudes such as 45 degrees, summer insolation can exceed winter insolation by a factor of four or more due to both longer days and higher sun angles. At the poles, insolation swings from zero during the polar night to values comparable to mid-latitude summers during the midnight-sun season, although the low angle still limits peak flux significantly. Earth's axial tilt of 23.44 degrees is the primary cause of this seasonal cycle, with orbital eccentricity modulating its amplitude between the two hemispheres. Latitude Radiation Budget Calculator works with annual means, so the seasonal cycle is averaged out.
What is the radiation budget and how does it differ by latitude?
The radiation budget at each latitude is the difference between absorbed solar radiation and emitted longwave infrared radiation at the top of the atmosphere. In the tropics, absorbed solar energy exceeds emitted infrared, creating an energy surplus of roughly 55 to 70 W/m2. In polar regions the opposite holds: emitted longwave radiation exceeds absorbed solar, creating a deficit that reaches about minus 100 to minus 120 W/m2. The atmosphere and ocean transport energy poleward to compensate. The ocean dominates the transport within roughly 15 degrees of the equator, peaking near 1.7 PW around 10 to 15 degrees north, while the atmosphere carries most of the flux poleward of about 30 degrees. This transport system is driven by the latitudinal radiation budget gradient and shapes global climate patterns including the jet streams and ocean gyres.
How does the polar radiation budget differ from the equatorial budget?
The equatorial radiation budget shows a persistent surplus throughout the year because high sun angles deliver intense solar energy while moderate surface temperatures limit longwave emission. Absorbed solar radiation near the equator averages around 315 to 325 W/m2 annually, against outgoing longwave near 250 to 270 W/m2. In contrast, polar regions receive little solar energy due to low angles and long winter darkness, yet their surfaces emit significant longwave radiation year-round because cold surfaces still radiate according to the Stefan-Boltzmann law. The resulting annual deficit at the poles exceeds minus 100 W/m2. Without continuous poleward heat transport from the tropics, polar temperatures would fall far below observed values.
At what latitude does the net radiation budget flip from surplus to deficit, and why does this model give 35.2 degrees?
In the two-term Legendre model used here, net radiation is a constant plus a term proportional to P2(x). If absorbed shortwave and outgoing longwave have the same global mean, the constant vanishes and the sign change occurs exactly where P2(x) = 0, that is where sin(latitude) = 1/sqrt(3) = 0.5774, or 35.26 degrees in each hemisphere. With the default settings the two global means differ slightly, by 0.12 W/m2, which shifts the crossing marginally to 35.2 degrees. Observed profiles put the crossing nearer 36 to 40 degrees because the real albedo and outgoing longwave profiles carry structure beyond a single Legendre mode: subtropical clear-sky regions absorb more sunlight than a pure P2 albedo allows, and outgoing longwave peaks in the subtropics rather than at the equator. The model therefore places the crossing a few degrees equatorward of observation and understates the near-equatorial surplus by roughly 10 percent, and it cannot reproduce the small equatorial dip in outgoing longwave caused by deep convective cloud.
How is poleward heat transport calculated from the radiation budget?
In a steady state with no local energy storage, the energy crossing a latitude circle must equal everything the surface area on one side of it gains from net radiation. Integrating the net top-of-atmosphere flux over the spherical cap from the south pole to latitude phi gives F(phi) = 2 pi R_E^2 times the integral from -1 to sin(phi) of the net radiation anomaly with respect to x. Because the anomaly here is a pure P2 term and the integral of P2 from -1 to x equals minus x(1 - x^2)/2, the transport collapses to the closed form F(phi) = pi R_E^2 (A0 B2 + L2) x (1 - x^2). This is maximised where the derivative of x(1 - x^2) vanishes, at x = 1/sqrt(3), giving a peak of pi R_E^2 (A0 B2 + L2) times 2/(3 sqrt 3) at 35.26 degrees. With the default settings that is about 5.4 PW, consistent with the 5 to 6 PW range inferred from satellite fluxes and atmospheric reanalysis.
How does the latitude radiation budget influence ocean and atmospheric circulation?
The latitude-dependent radiation budget is the fundamental engine driving large-scale atmospheric and oceanic circulation. The tropical radiation surplus heats air and ocean surface water, causing air to rise in the Intertropical Convergence Zone and ocean water to flow poleward in western boundary currents like the Gulf Stream and Kuroshio. The polar radiation deficit cools air and ocean, causing dense water to sink in high-latitude regions such as the North Atlantic Deep Water formation zone, driving the thermohaline circulation. In the atmosphere, the temperature gradient between tropics and poles maintains the jet streams and the Hadley, Ferrel, and Polar circulation cells. Changes to this radiation gradient due to polar amplification of warming can alter circulation patterns and weather extremes at mid-latitudes.
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Reviewed for accuracy by Daniel Agrici, Founder & Lead Developer ยท Editorial policy
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