找回密码
 马上注册

QQ登录

只需一步,快速开始

查看: 7351|回复: 3

电子差速器程序

[复制链接]
发表于 2014-10-8 20:04:57 | 显示全部楼层 |阅读模式
求电子差速器程序或者思想
如果您觉得我的帖子对您有用,请不吝给我一个“赞”!
发表于 2014-10-8 22:18:19 | 显示全部楼层
看一下The LEGO MINDSTORMS EV3 Laboratory,里边有详细说明
如果您觉得我的帖子对您有用,请不吝给我一个“赞”!
回复

使用道具 举报

发表于 2014-10-8 22:22:48 | 显示全部楼层
Let H be the distance between the rear and the
front wheels and D be half the distance between the rear
wheels. Let V be the car’s speed and a be the steering
angle. S is defined as the car’s turning radius. Using
the geometric relationships, and knowing that the angle
marked with a black square is a right angle (90°), we have
S = H / tan(a). The steering angle a must be converted into
radians: This is done by multiplying the angle in degrees
by (π / 180), which in this case comes to approximately
0.017. Also, for small values of angle a [rad], tan(a) can
be approximated with the value of a itself, so we can write
the approximate equation S ≈ H / (0.017 × a).
When the car is turning, W = V / S, where W is
the angular speed, V is the car’s speed, and S is the
turning radius. The speed of the outer wheel (the right
wheel in Figure 12-1) is VR = W × (S + D). Using some
substitutions and simple manipulations, we end up
with VR = V / S × (S + D) = V × (1 + D / S), which finally
yields VR = V × (1 + D × 0.017 × a / H). Similarly, for
the inner wheel (the left wheel in Figure 12-1), we get
VL = V × (1  D × 0.017 × a / H).
如果您觉得我的帖子对您有用,请不吝给我一个“赞”!
回复

使用道具 举报

 楼主| 发表于 2014-10-9 10:54:06 | 显示全部楼层
shiren-zelda 发表于 2014-10-8 22:22
Let H be the distance between the rear and the
front wheels and D be half the distance between the  ...

谢谢
如果您觉得我的帖子对您有用,请不吝给我一个“赞”!
回复

使用道具 举报

您需要登录后才可以回帖 登录 | 马上注册

本版积分规则

手机版|中文乐高 ( 桂ICP备13001575号-7 )

GMT+8, 2024-12-22 19:41 , Processed in 0.104431 second(s), 23 queries .

Powered by Discuz! X3.5

Copyright © 2001-2020, Tencent Cloud.

快速回复 返回顶部 返回列表