Remember that the Earth rotates at 15 degrees per hour, and 0.25 degrees per minute. It can get confusing when you're comparing spatial seconds with temporal seconds, right? That's why we will stick with decimal notation in all of our references to latitude ($\phi$), longitude ($\lambda $$$ ), and angles (degrees).

#### Self-Check

**Step 1: Correcting for Standard Time (Standard Meridian) in Hours: **Find the time zone of your client. The hourly longitude correction is plus or minus X hours from the Prime Meridian, where UTC = 0h. Multiply the positive or negative hour value by 15°, and you will know your standard meridian for the Time Zone.

*Key:* Your standard meridian for your time zone begins on the East side of the time zone. (Sun rises in the East, right?)

**Step 2: Correcting for Time from the Local Longitude Relative to Standard Meridian in Minutes: **

You need to know the longitude for the **standard meridian** of the client ( ${\lambda}_{std}$) from their local time zone, and the **local longitude** of the client's location in question (${\lambda}_{loc}$). We calculate the longitudinal correction, ${t}_{\lambda}$ from these two.

Keep in mind: Time zone borders are political boundaries, and can be constructed on both sides of a Standard Meridian. A locale to the east of the Standard Meridian would still be input as a positive value (the resulting minutes would be positive).

$${t}_{\lambda}=-4(\frac{min}{deg})({\lambda}_{std}-{\lambda}_{loc})[min]$$

Note: we use the sign convention that longitudes are *positive* valued for to the East of the Prime Meridian, and *negative* valued to the West of the Prime Meridian. This equation is valid for both sides of the Prime Meridian. The exterior negative sign is there to make the time correction algorithm work correctly.

**Step 3: Calculating Equation of Time (Analemma) Correction in Minutes **(*E _{t}*): We begin with the two simple coefficients of

*n*(the day of the year, from 1-365), and

*B*(see the first equation). Our assumption is that the year begins at midnight of the new year, and the trigonometric portions of the equation of time will take an argument of "B" in degrees.

$$B=(n-1)\frac{{360}^{\circ}}{365}$$

$${E}_{t}=229.2(0.000075)+229.2(0.001868\mathrm{cos}B-0.032077\mathrm{sin}B)-\phantom{\rule{0ex}{0ex}}229.2(0.014615\mathrm{cos}2B+0.04089\mathrm{sin}2B)[min]$$

You have seen that the Equation of Time has a graphical representation, the *analemma*. Once we determine a correction in the scale of minutes, we can use it in the Time Correction Factor, *TC*.

**Step 4: Completing the Time Correction Factor **($TC$): The time correction factor is a time shift in minutes.

$TC={t}_{\lambda}+{E}_{t}$

Recall that 0.25° of longitudinal rotation (of the planet) will consume 1 minute of time. That would make each degree of change equivalent to four minutes. Hence, we need to multiply our longitudinal correction by a factor of four $$\frac{\mathrm{min}}{\mathrm{deg}}$$ to yield a consistent unit of minutes in time. This is shown as the value of the longitudinal correction (*L _{c}*) in units of minutes (temporal, not geospatial).

**Step 5: Accounting for Daylight Savings Time (DST) in 60 minutes.**

The equations do not tell you when DST occurs from country to country. There is a +60 minute difference between March and November in the USA. So, if we had to correct for DST, then we would need to *subtract* that 60 minute addition back out.

$$Local\text{}Solar\text{}Time\text{}(tsol)=\{\begin{array}{c}Standard\text{}Time+TC\text{ifStandardtime.}\\ Standard\text{}Time\text{}+\text{}TC\text{}-\text{}60\text{ifDaylightSavingstime.}\end{array}$$

Again, make sure that the data is presented in minutes, rather than hours.