In reading "The Error in Kepler's Acronychal Data for Mars" by Curtis Wilson 1969 (unfortunately , this is not an open article), we read the following paragraph:

How did it come about that Tycho’s solar theory was so inaccurate, yielding errors over 7’? The main source of error is Tycho’s assumption of 3’ as the horizontal parallax of the sun, which implies a parallax of 2’30” at an altitude of 34”5’, the noonday altitude of the sun at Uraniburg at the time of the equinoxes; the correct value would be 7”. An additional source is Tycho’s table of refractions, which gives 45“ at an altitude of 34”5’, some 40” too small, Now the altitudes of the sun at or near the time of the equinoxes were the data most strongly determining the eccentricity, since the sun was then about a quadrant’s distance from aphelion. Tycho’s two errors cause him to add 1’45” to the apparent altitude to obtain the true altitude, whereas about 1’18” should be subtracted. The total difference of 3’ between Tycho’s value and the sun’s true altitude implies a displacement of the sun in longitude by over 7‘.

I don't see how 3’ in altitude amount to 7‘ in longitude (in the ecliptic coordinate system I'm convinced is meant here; or I miss something). I would rather say that 3’ mistake in altitude results in 3’ in longitude at max - and usually even less. From what I was able to see this derivation of 7‘ in the article is not explained elsewhere in the paper.


re-reading the paper, I think it is necessary to add more context, as I'm afraid the 7‘ is not computed but only given. In the article it is said that in the model of the Sun that Tycho Brahe developed the eccentricity is too high [Which is indeed true - about X2], and this results in 7‘ discrepancy between the real location of the Sun and the model-location; this +7‘ in the early spring and -7‘ in the fall.

But solar theories prior to the late 17th century erred in giving the sun's (or earth's) orbit an exaggerated eccentricity. In the case of the Tychonic theory that Kepler used, this has the effect of putting the earth about 7' ahead of its true position at the beginning of spring.

I'm still lost as how a 3min error in altitude measurement can produce so an erroneous model. Well, there is the option of the model itself that was wrong: namely an eccentric with equant as center versus the real model of ellipse. But I have reason to believe this is not enough.


1 Answer 1


I think I have it now. Basically I was thinking the other way around: the mode of thinking was: "What longitude component will arise from 3 arc min in altitude?" Indeed, if, for example, Jupiter will be observed 3 arc min higher than it really is, it will contribute to the longitude component no more than 3 arc minutes. The 3 minutes angular distance will be splited between longitude and latitude components. But the question I should have asked is rather: "how much we have to move in longitude, while keeping the same latitude, in order the result to be 3 arc min higher in altitude?". This is the question to be asked because it was given that the Sun keeps the same latitude by definition.

The answer to this question is not constant across time of day or equally across the year. Following what the paper suggest, lets see what happens at noon time at the equinoxes. When the at the meridian, the declination line is parallel to the horizon, hence, at North Hemisphere, if we want to see an object 3’ higher when at the meridian (South), it should have exactly 3’ higher declination. So our question is down to this: "How much we need to move on the ecliptic from the March/September equinox, so that the declination will be 3’ higher". First, This depends at which equinox we are at, For at the March equinox we should go forward (coming closer to Summer/June) - hence increase longitude; While at September we should go back-wards (again - coming closer to Summer/June) - hence decrease longitude. Let's denote the degrees that we should with $x$, then since the angle between the planes of the equator and ecliptic is $23.5$ then: $3’/23.5° = sin(x)$ or $0.05°/23.5° = sin(x)$. We have $x = 0.1219°$ or $x = 7’ 19”$.

So, indeed at Spring we moved forward the sun by more than 7’, and at Fall we moved the Sun backwards at the same amount.

  • $\begingroup$ I have taken: $lat =\sin(x)*23.5$ . This is indeed good approximation on those angles. But more accurate mathematical formulation (from spherical sine rule) is: $\sin(lat) = sin(x)*sin(23.5)$. So if we put lat=3 min, we would get 0.125394226, or 7min 31sec $\endgroup$
    – d_e
    Nov 5, 2021 at 7:56

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