攻擊距離的斜角算法? - 仙境傳說
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By Quintina
at 2017-02-16T00:02
at 2017-02-16T00:02
Table of Contents
鎖定型技能都會有攻擊距離格數
但當施放者與對象為相對斜角時
該怎麼算呢?
若攻擊距離6格
施展者位座標(0,0)
可以命中(6,0)與(0,6)
這是單純的水平與垂直距離
經測試一度以為是:
X+Y軸≦攻擊距離 即可命中
但發現(1,6) (6,1)等座標也ok
明顯非X軸值+Y軸值的簡略算法
也測試攻擊距離10格的技能
即施展者位(0,0)
可命中水平(10,0)與垂直(0,10)對象
而當對象位(10,3) (3,10)等座標時
也能成功命中
到底是怎麼計算的呢?
文組請求各位支援Q_Q
--
但當施放者與對象為相對斜角時
該怎麼算呢?
若攻擊距離6格
施展者位座標(0,0)
可以命中(6,0)與(0,6)
這是單純的水平與垂直距離
經測試一度以為是:
X+Y軸≦攻擊距離 即可命中
但發現(1,6) (6,1)等座標也ok
明顯非X軸值+Y軸值的簡略算法
也測試攻擊距離10格的技能
即施展者位(0,0)
可命中水平(10,0)與垂直(0,10)對象
而當對象位(10,3) (3,10)等座標時
也能成功命中
到底是怎麼計算的呢?
文組請求各位支援Q_Q
--
Tags:
仙境傳說
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