Javascript Typescript Omit 似乎将联合转变为交集

我有这种类型

type Cartesian = { kind: 'cartesian'; x: number; y: number; }
type Polar = { kind: 'polar'; angle: number; distance: number }
type Movement = { name: string } & (Cartesian | Polar);

我可以这样使用

const move = (movement: Movement) => { /* whatever */ };
move({ name: 'top right', kind: 'cartesian', x: 10, y: 10 });
move({ name: 'left', kind: 'polar', angle: Math.PI, distance: 10 });

但出于某种原因，我不能这样使用它

const unnamedMove = (unnamedMovement: Omit<Movement, 'name'>) => {
    move({ name: 'default', ...unnamedMovement })
}

因为TS抛出2345:

Argument of type '{ kind: "cartesian" | "polar"; name: string; }' is not assignable to parameter of type 'Movement'.
  Type '{ kind: "cartesian" | "polar"; name: string; }' is not assignable to type '{ name: string; } & Polar'.
    Type '{ kind: "cartesian" | "polar"; name: string; }' is missing the following properties from type 'Polar': angle, distance

我不明白这一点.

如果我没有弄错的话，Omit<Movement, 'name'>应该相当于Cartesian | Polar型的联合体，这将使{ name: 'default', ...unnamedMovement }成为Movement，所有的东西都应该工作.

然而，看起来TS推断Omit<Movement, 'name'>就像它是联合类型Cartesian & Polar一样，因此是错误的.

这是我的错误还是错误？

playground link