Nonrecursive Movement Formulas/zh

由於算術幾何序列有明確的公式，我們可以建立非遞歸函數來計算簡單但有用的結果，例如在任一 tick 上玩家的高度，或者在初始速度與持續時間的基礎上計算跳躍的距離。

定義：

• ${\textstyle v_{0}}$ 是玩家的初始速度（跳躍之前，$t_{0}$ 時的速度）
• ${\textstyle t}$ 是計入的 ticks 數（例如：平地上 t=12，參見跳躍持續時間）
• ${\textstyle J}$ 是「跳躍增益」（疾跑跳躍為 0.3274，斜疾跑跳躍為 0.291924，45°無疾跑跳躍為 1.0……）
• ${\textstyle M}$ 是跳躍後的移動乘數（45°斜疾跑為 1.3，正常疾跑為 1.274，無疾跑45°為 1.0……）

垂直運動（跳躍）[1.8]

跳躍後的垂直速度（$t\geq 6$）

${\textstyle {\textrm {V}}_{Y}(t)=4\times 0.98^{t-5}-3.92}$

跳躍後的相對高度（$t\geq 6$）

Failed to parse (syntax error): {\textstyle \textrm{Y}_{rel}(t) = \underset{\textrm{跳跃最高点}}{\underbrace{197.4 - 217 \times 0.98^5}} + 200 (0.98-0.98^{t-4}) - 3.92 (t-5)}

對於 ${\textstyle t<6}$ 的情況，見下文。

垂直運動（跳躍）[1.9+]

跳躍後的垂直速度（$t\geq 1$）

${\textstyle {\textrm {V}}_{Y}(t)=0.42\times 0.98^{t-1}+4\times 0.98^{t}-3.92}$

跳躍後的相對高度（$t\geq 0$）

${\textstyle {\textrm {Y}}_{rel}(t)=217\times (1-0.98^{t})-3.92t}$

Horizontal Movement (instant jump)

假設玩家在跳躍前已經在空中。

疾跑跳躍後的水平速度（$t\geq 2$）

${\textstyle {\textrm {V}}_{H}(v_{0},t)={\frac {0.02M}{0.09}}+0.6\times 0.91^{t}\times \left(v_{0}+{\frac {J}{0.91}}-{\frac {0.02M}{0.6\times 0.91\times 0.09}}\right)}$

疾跑跳躍距離（$t\geq 2$）

${\textstyle {\textrm {Dist}}(v_{0},t)=1.91v_{0}+J+{\frac {0.02M}{0.09}}(t-2)+{\frac {0.6\times 0.91^{2}}{0.09}}\times (1-0.91^{t-2})\times \left(v_{0}+{\frac {J}{0.91}}-{\frac {0.02M}{0.6\times 0.91\times 0.09}}\right)}$

Note: These formulas are accurate for most values of $v_{0}$, but some negative values can wind up activating the speed threshold and reset the player's speed at some point, thus rendering these formulas inaccurate.

Horizontal Movement (delayed jump)

Assuming the player is on ground before jumping (at least 1 tick since landing).

Horizontal speed after sprintjumping ($t\geq 2$)

${\textstyle {\textrm {V}}_{H}^{*}(v_{0},t)={\frac {0.02M}{0.09}}+0.6\times 0.91^{t}\times \left(0.6v_{0}+{\frac {J}{0.91}}-{\frac {0.02M}{0.6\times 0.91\times 0.09}}\right)}$

Sprintjump distance ($t\geq 2$)

${\textstyle {\textrm {Dist}}^{*}(v_{0},t)=1.546v_{0}+J+{\frac {0.02M}{0.09}}(t-2)+{\frac {0.6\times 0.91^{2}}{0.09}}\times (1-0.91^{t-2})\times \left(0.6v_{0}+{\frac {J}{0.91}}-{\frac {0.02M}{0.6\times 0.91\times 0.09}}\right)}$

Advanced Formulas

Horizontal speed after $n$ consecutive sprintjumps on a momentum of period ${\textstyle T}$ ($n\geq 0$, $T\geq 2$).

${\textstyle {\textrm {V}}_{H}^{\,n}(v_{0},T,n)=\left(0.6\times 0.91^{T}\right)^{n}v_{0}+\left(0.6\times 0.91^{T-1}J+0.02M{\frac {1-0.91^{T-1}}{0.09}}\right){\frac {1-(0.6\times 0.91^{T})^{n}}{1-0.6\times 0.91^{T}}}}$

If the first sprintjump is delayed, multiply ${\textstyle v_{0}}$ by 0.6