Before you can read the parent note, you must own every symbol it throws at you. This page builds each one from nothing — plain words, then a picture, then why the topic needs it. Each block leans on the one before it.
Why the topic needs this: a plane change changes the direction of the arrow while (often) keeping its length; a burn to circularize changes its length. If you only think of velocity as a number, you literally cannot see the difference between these two — you need the arrow.
What does subtracting arrows look like? Lay v1 and v2tail-to-tail (start them from the same point). The arrow drawn from the tip of v1 to the tip of v2 is Δv — the missing side that closes the triangle.
Picture: θ=0 means the two arrows point the same way (no turn, only a possible length change). θ=90∘ means the new arrow points sideways to the old one. Larger θ = a sharper turn = a longer Δv arrow to bridge the gap.
Why the topic needs it: θ (and its total i) is the dial you are optimizing. The whole "optimal split" question is: given a required total tilt i, how much of that turn do you do at burn 1 versus the rest at burn 2? See Plane Change Maneuvers.
Why the topic needs it: it is the algebra that derives the Law of Cosines for our arrows. To get the length of Δv we dot it with itself:
Δv2=Δv⋅Δv=(v2−v1)⋅(v2−v1)=v22−2v1⋅v2+v12.
Then v1⋅v2=v1v2cosθ turns the middle term into an angle — reproducing exactly the boxed formula. The dot product is the machine that converts arrows into that formula.
Now take the special case where the burn does not change the speed, only the direction. That means the before and after arrows have the same length: v1=v2=v. This is the crucial assumption — drop it and the tidy formula below no longer holds.
Substitute v1=v2=v into the boxed formula:
Δv=v2+v2−2vvcosθ=v2−2cosθ.
Where does the half-angle come from? Because both sides are equal length v, the velocity triangle is isosceles (two equal sides). Drop a straight line from the tip down the middle: it splits the tip-angle θ into two equal halves of θ/2, and splits the base (Δv) into two equal halves — that even split is forced by the two sides being equal, not merely convenient. Look at figure s05: each half is a right triangle whose hypotenuse is v and whose "opposite" side is 21Δv. By the definition of sine,
sin(2θ)=v21Δv⟹Δv=2vsin(2θ)
Check: at θ=0, sin0=0, so Δv=0 — no turn, no cost. As θ grows the cost grows, and crucially it is proportional to v: the faster you go, the more a turn costs.
Why the topic needs it: the pure-turn cost 2vsin(θ/2) is proportional to the speed v at the moment you turn. So turning at the fast near-point is brutal, turning at the slow far-point is cheap. This single fact is the reason the optimal split shoves nearly all the plane change out to apoapsis. The transfer between two circular orbits that sets up these fast/slow burns is the Hohmann Transfer Orbit; pushing the far point even further out to turn even more cheaply is the Bi-elliptic Transfer.
Why the topic needs it: plot the total cost Δvtot(s) against the split s and it dips to a lowest point — the cheapest split. That lowest point is exactly where the slope is zero, which is why the parent sets dsdΔvtot=0. Reading that condition out loud gives "the marginal cost of one more degree of turn is equal at both burns" — shift a degree toward whichever burn is cheaper until they balance.
