True anomaly from eccentric anomaly: Visual walkthrough

Child of True anomaly from eccentric anomaly. Here we redraw the whole conversion from zero. Every step gets one picture. Nothing is assumed — if we use a symbol, you'll have seen it built.

Step 0 — What are we even looking at?

Before any formula, let's earn every word.

Definition The cast of characters (all pictured below)

An ellipse — a squashed circle. It's the path the planet actually travels.
The center $O$ — the middle of the ellipse, where the two axes cross.
The focus $F$ — a special off-center point. The Sun sits here. This is not the center.
Perihelion — the point on the orbit closest to the focus. We measure all angles starting from here.
The planet at point $P$ .
$a$ = semi-major axis = half the long width of the ellipse.
$e$ = eccentricity = "how squashed" — a number from $0$ (perfect circle) up toward $1$ (very stretched). Built in Orbit geometry — semi-major axis and eccentricity.
$c$ = focal distance = the gap from center $O$ to focus $F$ . We define $c : = a e$ (center-to-focus distance). This is the number we subtract when we shift viewpoint in Step 3.

Look at the picture. The teal dot is the center $O$ . The orange dot to its right is the focus $F$ — notice it is pushed off the center toward perihelion by exactly $c = a e$ .

We want the angle as seen from the orange focus — that is the true anomaly $ν$ . But the easy angle to compute (from Kepler's Equation, Kepler's Equation) is measured from the teal center — that is the eccentric anomaly $E$ . The whole page is: given $E$ , produce $ν$ .

Step 1 — Draw the helper circle and the angle $E$

Here is how $E$ is defined, precisely. Take the planet $P$ on the ellipse. Go straight up (or down) until you hit the auxiliary circle at a point $P^{'}$ . The angle from the center $O$ , measured from perihelion to $P^{'}$ , is the ==eccentric anomaly $E$ ==.

Step 2 — Squash the circle to land on the ellipse

Look at the vertical dotted line: $P^{'}$ on the circle drops down to $P$ on the ellipse. The $x$ stays the same, the $y$ gets multiplied by $b / a$ .

Step 3 — Slide the origin over to the focus

The focus sits a distance $c = a e$ toward perihelion — same $e$ as in Orbit geometry — semi-major axis and eccentricity. Bigger $e$ = focus further from center.

The orange arrow in the figure is the planet's position vector from the focus. Its horizontal part is the old $x_{O}$ minus the offset $c = a e$ :

Step 4 — Read off the distance $r$

Pythagoras on the orange right triangle: $r^{2} = x_{F}^{2} + y_{F}^{2}$ .

Step 5 — Turn the arrow into the angle $ν$

Step 6 — Fold both into one clean formula (the half-angle gem)

Build $1 \pm cos ν$ over a common denominator: $1 + cos ν = 1 + \frac{c o s E - e}{1 - e c o s E} = \frac{( 1 - e c o s E ) + ( c o s E - e )}{1 - e c o s E} = \frac{( 1 - e ) ( 1 + c o s E )}{1 - e c o s E}$ $1 - cos ν = \frac{( 1 - e c o s E ) - ( c o s E - e )}{1 - e c o s E} = \frac{( 1 + e ) ( 1 - c o s E )}{1 - e c o s E}$ Divide — the ugly denominator $1 - e cos E$ cancels: $\frac{1 - c o s ν}{1 + c o s ν} = \frac{( 1 + e ) ( 1 - c o s E )}{( 1 - e ) ( 1 + c o s E )}$ Apply the half-angle identity to both sides: $tan^{2} \frac{ν}{2} = \frac{1 + e}{1 - e} tan^{2} \frac{E}{2}$

Common mistake The half-angle formula is monotone, but inverting it needs care at

$E = π$ It's tempting to say "just apply $arctan$ to both sides." But $tan \frac{E}{2}$ has a pole (blows up to $+ \infty$ ) at $E = π$ , and $arctan$ only returns values in $(- \frac{π}{2}, \frac{π}{2})$ — i.e. it gives $\frac{ν}{2}$ in one narrow branch. Feed it $tan \frac{E}{2}$ raw and, as $E$ crosses $π$ , $arctan$ wraps back, dropping you onto the wrong branch and putting $ν$ in the wrong revolution. The clean fix: never invert the ratio directly. Instead form the two half-angle pieces separately, $tan \frac{ν}{2} = \frac{1 + e s i n \frac{E}{2}}{1 - e c o s \frac{E}{2}},$ and feed the numerator and denominator into atan2: $\frac{ν}{2} = atan2 (1 + e sin \frac{E}{2}, 1 - e cos \frac{E}{2}), ν = 2 \cdot \frac{ν}{2} .$ Because atan2 reads the sign of each piece, it tracks $\frac{ν}{2}$ smoothly through the $E = π$ pole across the full orbit — the monotone curve below is exactly this correctly-branched result.

The figure plots $ν$ vs $E$ for several $e$ , computed with the atan2 recipe above. It's a single smooth increasing curve each time — that monotonicity (once the branch is chosen correctly) is the whole reason we prefer this form.

Step 7 — The trap: why $arccos$ alone fails (degenerate viewpoint)

The two orange dots have the same $cos ν$ but opposite $sin ν$ . $arccos$ can't tell them apart.

Step 8 — Edge cases, all of them

The one-picture summary

One frame carries the whole chain: circle → squash to ellipse → slide origin to focus (by $c = a e$ ) → read the angle. The teal machinery is the center world (where $E$ lives); the orange machinery is the focus world (where $ν$ lives). Everything else is Pythagoras and one half-angle identity.

Recall Feynman retelling — the walkthrough in plain words

We drew a perfect circle hugging the oval orbit and marked a helper angle $E$ from the middle. Sliding straight down from the circle lands us on the oval — that's how the helper angle points to the real planet. But the Sun doesn't sit in the middle; it sits off to one side, at the focus, a distance $c = a e$ away. So we picked up our whole coordinate system and slid it over to the Sun. From there we measured a right triangle: across, up, and the slanted distance $r$ to the planet. "Across over slant" gave $cos ν$ ; "up over slant" gave $sin ν$ . Both together pin the real angle $ν$ without any guessing. To make it one neat recipe, we used a half-angle trick and out fell $tan \frac{ν}{2} = \frac{1 + e}{1 - e} tan \frac{E}{2}$ . The square-root factor is the "Sun-is-off-center" stretch: near the Sun a small helper step swings a big real angle. One warning: don't blindly $arctan$ that ratio — at aphelion it blows up, so hand the top and bottom to atan2 and it stays smooth. Round orbit? No offset, no stretch, all three angles are the same.

Recall

Give $r$ in terms of $E$ . ::: $r = a (1 - e cos E)$ Why take the positive square root for $r$ ? ::: $r$ is a distance so $r \geq 0$ ; and $a (1 - e cos E) > 0$ for a valid ellipse, so the negative root is physically meaningless. Why must we shift origin to the focus, and by how much? ::: $ν$ is measured at the focus; from the center you'd get $E$ instead. Shift by the focal distance $c = a e$ . Why prefer the half-angle $tan$ formula, and what's the catch when inverting? ::: $tan (θ /2)$ is monotone over a full orbit — no quadrant ambiguity, unlike $arccos$ . Catch: $tan (E /2)$ has a pole at $E = π$ , so invert with atan2 on the separate numerator/denominator, not a raw $arctan$ . What is the stretch factor and its $e = 0$ value? ::: $(1 + e) / (1 - e)$ ; equals $1$ when $e = 0$ so $ν = E$ . Why the $+$ branch of the square root? ::: The stretch factor is positive, so $tan (ν /2)$ and $tan (E /2)$ share the same sign — $ν$ and $E$ lie in the same half of the orbit.

See also: Position and velocity in the perifocal frame · Orbital radius equation · Kepler's Equation.