Measuring the elongation of Venus with a watch?

Updated 3 May 26

As an alternative, I thought instead about just using the timings of the settings of the Sun and of Venus (I'm not a morning person!). Thinking that no doubt someone has already investigated this, after some digging, I came across this site with a method for such a measurement. However, there's no derivation of the method and it seems to assume that both will set at the same azimuth, which doesn't seem right. Anyway, I thought I'd check it with Stellarium and chose the greatest elongation East for Venus of 15 August 2026:

• Sun and Venus are about 45° apart, as we'd expect
• the Sun's declination (angular distance above the celestial equator) is around 14°
• that of Venus is around -5° (i.e. it's below the celestial equator), so very different from that of the Sun
• the Sun straddles the horizon, i.e. is half set at around 20.20 (BST?) at azimuth 294°
• Venus sets about 80 minutes later, at azimuth 261°, i.e. there's a whopping 33° difference!

If we use the formula, it says for a 45° separation, Venus sets 3 hours, not 1 hour, after the Sun, so the formula is indeed wrong - the Sun and Venus will set at the same point on the horizon only if they are at the same declination, which is not the case here nor in general. As a comparison, the star Denebola *is* at about the same declination as the Sun at this time, with an angular separation from it of 31°. The formula gives about 2 hours between the Sun and Denebola setting and Stellarium confirms that.

In addition to the time between the setting of the Sun and of Venus, then we'd need the difference in azimuths at setting or the declinations - and a new formula.

Verdict: What's the point?