IAO 2013, Теория, Група β — Задача 4. XVIII century. Midday.

Solar constant in historical units

Horse-powers per square verst is a power per unit of area, the quantity that dimension is similar to the solar constant. In physics, it is properly called the “power flux”.

Let us calculate the solar constant in units, which were coming into operation in the end of the XVIII century.

I method — visual.

The energy-release of the Sun has not changed since the end of the XVIII century and in units of the horse-power is

The distance from the Sun to Earth has not changed since the end of the XVIII century as well, and in versts is

Thus, the solar constant in units of the end of the XVIII century is

II method — formal.

Source note: the key’s subscript “XIX” is retained although the problem concerns the XVIII century.

4.1. Midday power flux

The outskirts of Dubingiai in the XVIII century were located on the same latitude as where they are located in the XXI century, that is, at about latitudes 54.7°–55.2° (latitudes of Vilnius and Moletai respectively, may be found from the map). The value of 23.5°, the inclination of the equatorial plane of the ecliptic, also almost has not changed (strictly speaking, it had declined by about one arc minute). Thus, at the middays of the winter solstice in the XVIII century in the outskirts of the town Dubingiai the sun was at an altitude about

at the middays of spring and autumn — at an altitude about

at the middays of the summer solstice — at an altitude about

While the altitude of the Sun is h, the power flux to the unit of the territory is

Thus, the target values are:

• in winter — $W_W=A\sin h_W\approx420\,000\ \mathrm{hp/vrst^2}$;
• in spring and autumn — $W_0=A\sin h_0\approx1\,220\,000\ \mathrm{hp/vrst^2}$;
• in summer — $W_S=A\sin h_S\approx1\,810\,000\ \mathrm{hp/vrst^2}$.

4.2. Solar power incident on a horse

And what is the capacity of the solar energy incident at those times on a local horse? This value largely depends on the orientation of the horse relative to the Sun than the altitude of the sun above the horizon. Let us assume that the cross-sectional area of a horse perpendicular to the sun-rays is 1–3 arsh² (depending on the mentioned above orientation of the horse) and recalculate the value of the solar constant into horse-powers per square arshin.

Distance from the Sun

The solar constant in these units equals to

Thus, a horse standing in the sun, receives about one to three horse-powers of solar radiation! Just this can be a surprise.

Answer: 1–3 hp.

Note. In solving the problem we emphasize very often that many of the physical parameters are not changed in the last two centuries: the solar constant, the distance from Earth to the Sun, the latitude of location, inclination of the ecliptic to the equatorial plane. Students who also record this constancy, should be encouraged to add points.

Source note: the distance conversion prints $2.103\cdot10^{14}$ arsh, but the following formula uses $2.103\cdot10^{11}$ arsh. Both printed exponents are retained.

Official basic criteria

4.common. Totally 2 pt, including:

• Calculation of the solar constant in necessary units, result $2.12\cdot10^6$ hp/vrst² or 0.94 hp/arsh² — 2 pt.

4.1. Totally 3 pt, including:

• Latitudes the same as now (about 54.7°–55.2°) — 1 pt.
• Calculation of altitudes of midday Sun in winter, spring, autumn, summer — 1 pt.
• $W=A\sin h$, correct final calculations — 1 pt.

4.2. Totally 3 pt, including:

• Correct understanding on what the answer may depend — 0.5 pt.
• Correct model of horse, including the picture — 0.5 pt.
• The cross-sectional area of a horse perpendicular to the sun-rays is 1÷3 arsh² — 0.5 pt.
• Correct calculations and result about 1–3 hp — 1 pt.
• Common conclusion about surprise that a horse standing in the sun, receives about one to three horse-powers of solar radiation — 0.5 pt.