Nonrecursive Movement Formulas/zh

由于算术几何序列有明确的公式，我们可以建立非递归函数来计算简单但有用的结果，例如在任一 tick 上玩家的高度，或者在初始速度与持续时间的基础上计算跳跃的距离。

定义：

• ${\textstyle v_{0}}$ 是玩家的初始速度（跳跃之前，$t_{0}$ 时的速度）
• ${\textstyle t}$ 是计入的 ticks 数（例如：平地上 t=12，参见跳跃持续时间）
• ${\textstyle J}$ 是“跳跃增益”（疾跑跳跃为 0.3274，斜疾跑跳跃为 0.291924，45°无疾跑跳跃为 1.0……）
• ${\textstyle M}$ 是跳跃后的移动乘数（45°斜疾跑为 1.3，正常疾跑为 1.274，无疾跑45°为 1.0……）

垂直运动（跳跃）[1.8]

跳跃后的垂直速度（$t\geq 6$）

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

跳跃后的相对高度（$t\geq 6$）

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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)}

对于 ${\textstyle t<6}$ 的情况，见下文。

垂直运动（跳跃）[1.9+]

跳跃后的垂直速度（$t\geq 1$）

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

跳跃后的相对高度（$t\geq 0$）

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

Horizontal Movement (instant jump)

假设玩家在跳跃前已经在空中。

疾跑跳跃后的水平速度（$t\geq 2$）

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

疾跑跳跃距离（$t\geq 2$）

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

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