Fw: [認真] 要如何才能射的遠?
※ [本文轉錄自 sex 看板 #1JIM1Q6n ]
作者: aibltjv (乖小孩) 看板: sex
標題: Re: [認真] 要如何才能射的遠?
時間: Sat Apr 12 23:48:40 2014
小弟是認真發問的,魯弟的射精是用滴的,
女友也很不給面子說我的很廢,有沒有辦法讓他用噴的,
有聽說過有人可以射出20cm距離,怎麼精進呢?
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1.首先要先養精蓄銳,使用過量的話自然力量不大射不遠
2.在養精蓄銳之後,再開始考量射遠的技巧
先假設:
(1)目標地面和你所站的位置沒有高低差(亦即不考慮站在椅子上之類的)
(2)精液視為質點,無形變和阻力問題
(3)無其他外力(例如風力)
(4)你的腳到你的發射點高度為1公尺
現在你有一定的力道(動量)和一定的發射物質量(可決定初速v),要決定發射角度
水平方向速度 = vcosθ
垂直方向速度 = vsinθ- 9.8t(向上)
將速度積分,得垂直方向位置 = vsinθt - 4.9t^2 + c
由於初始位置在1
由此判定c = 1
亦即位置對時間函數 = vsinθt - 4.9t^2 + 1
設射出物落地經歷時間t'
即 0 = 4.9t'^2 - vsinθt' - 1
t' 約等於 0.102(vsinθ + √[(vsinθ)^2+19.6] )
時間乘上水平速度即是發射距離
亦即求出vcosθ*0.102(vsinθ + √[(vsinθ)^2+19.6] )的最大值
(注意:此處的v為已知)
假設粗估 v = 1(公尺/秒)好了
這個距離對發射角度的函數就變成
f(θ) = 0.102cosθsinθ + 0.102cosθ√[(sinθ)^2+19.6]
求極大值--> df(θ)/dθ = 0時有極大值(此考慮 0 < θ < π/4)
0 = 2cos2θ - sinθ√[(sinθ)^2+19.6] + sinθ(cosθ)^2/√[(sinθ)^2+19.6]
= 2cos2θ - (19.6sinθ-cos2θsinθ)/√[(sinθ)^2+19.6]
變數變換令sinθ = x,cos2θ = 1-2(sinθ)^2 = 1-2x^2
原式可化簡成:
0 = 2 - 4x^2 - (18.6x + 2x^3)/√(x^2 + 19.6)
移向、平方得
(4x^2 - 2)^2 = (18.6x + 2x^3)^2 / x^2 + 19.6
(x^2 + 19.6)[(4x^2 - 2)^2] = (18.6x + 2x^3)^2
估算(每一項四捨五入)得
g(x) = 12x^6 + 223.2x^4 - 655^2 + 78 = 0
再變數變換令y = x^2
h(y) = 12y^3 + 223.2y^2 - 655y + 78 = 0
此時就0 < y( = (sinθ)^2 ) < 0.5
利用勘根法,h(0.1)>0;h(0.2)<0
故y落在0.1到0.2之間
同理解小數第二位
h(0.12)約為2.6;h(0.13)約為-3.4
故y落在0.12到1.3之間
藉由線性內插估計y約為0.1243
即sinθ=√y約為0.3526
接著可以利用sin18度約為 0.309 和sin22.5度約為 0.386做線性內插
得到θ約為20.54度
結論:在上述前提下可得仰角 = 20度上下會有最遠射程
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當然現實考量下
你的發射點高度不可能恰好等於1公尺(不過相去不遠)
發射初速也不會等於每秒1公尺(不過也相去不遠)
你的發射物更不可能不受阻力黏滯力等複雜因素影響
甚至是它還會在空中還會邊飛邊蒸發
所以要很精準地說如何射遠是很難用理論解決的
結論是你就多嘗試
自然而然就會知道對自己最合適的發射角度
不經一事,不長一智
經驗自然會造就你更遠的射程
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恩而且如果身高越高,仰角要越低
如果出發點和地平面高度一致(類似坐在地上射精)
那45度是最佳角度
隨著出發點上升,仰角要越平
至於如果要射到比出發點高的地方
仰角就要大於4度
但現實是比較不會有這種情況
噓
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沒辦法...我不知道空氣阻力的參數量值
或著你可以稍稍降低g的值會比較準一些
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※ 編輯: aibltjv (140.112.245.138), 04/14/2014 18:26:06
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※ 發信站: 批踢踢實業坊(ptt.cc)
※ 轉錄者: iskandar (118.169.203.245), 04/14/2014 22:06:03
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