





Fig. 4–2.A reversible machine. 一个可逆的机器
Suppose we have a reversible machine[A1] which is going to lift this distance X , three for one. We set up three balls in a rack which does not move,as shown in Fig. 4–2. Oneball is held on a stage at a distance one foot above the ground. The machinecan lift three balls, lowering one by a distance 1 . Now, we have arranged that the platform which holds three balls hasa floor and two shelves, exactly spaced at distance X , and further, that the rack which holds the balls is spaced atdistance X , (a). First we roll the balls horizontally from the rack to theshelves, (b), and we suppose that this takes no energy because we do notchange the height. The reversible machine then operates: it lowers the singleball to the floor, and it lifts the rack a distance X , (c). Now we have ingeniously arranged the rack so that theseballs are again even with the platforms. Thus we unload the balls onto therack, (d); having unloaded the balls, we can restore the machine to itsoriginal condition. Now we have three balls on the upper three shelves and oneat the bottom. But the strange thing is that, in a certain way of speaking, wehave not lifted two of them at all because, after all, there were ballson shelves 2 and 3 before. The resulting effect has been to lift one ball adistance 3X . Now, if 3X exceeds one foot, then we can lower the ball to return themachine to the initial condition, (f), and we can run the apparatus again.Therefore 3X cannot exceed one foot, for if 3X exceeds one foot we can make perpetual motion. Likewise, we can provethat one foot cannot exceed 3X , by making the whole machine run the opposite way, since it is areversible machine. Therefore 3X is neither greater nor less than a foot, and we discover then,by argument alone, the law that X=13 foot. The generalization is clear: one pound falls a certaindistance in operating a reversible machine; then the machine can lift p pounds this distance divided by p . Another way of putting the result is that three pounds times theheight lifted, which in our problem was X , is equal to one pound times the distance lowered, which is one footin this case. If we take all the weights and multiply them by the heights atwhich they are now, above the floor, let the machine operate, and then multiplyall the weights by all the heights again, there will be no change. (Wehave to generalize the example where we moved only one weight to the case wherewhen we lower one we lift several different ones—but that is easy.)
假设我们有一个可反转的机器,它可以通过降低三个X的距离,来抬高一个X的距离。我们在一个固定的支架上,放三个球,如图4-2。另一个球,在一定距离之外被托着,比地面高一英尺。此机器,可以通过把一个球降低1英尺,来抬高三个球。现在,有个装三个球的平台,它有三层,每层高都是X,我们把它放在距支架X处,(a)。首先,我们把球,滚到平台的各层上,(b),我们假定这不消耗能量,因为我们并未改变高度。然后,可反转机器开始运行:它把那个单个的球,降到地板,从而把平台抬高一个X的距离,(c)。现在,我们巧妙地安排平台,已让这些球,与架子相平。这样,我们就把这些球,卸载到了架子上,(d);卸载了这些球之后,我们就可以把机器,恢复到起原始状态。现在三个球在架子的上三层,一个在底层。但是,以某种方式来说,奇怪的事情就是,我们根本没有抬高其中的两个球,因为,毕竟这两个球,以前就是在第二层和第三层。实际结果就是,把一个球,抬高了3X的距离。现在,如果3X超过了1英尺,那么,我们就可以把球降低,然后让机器回到它的初始状态,(f),这样,我们就可以重新运行这个装置了。因此,3X不能超过1英尺,因为如果那样,等于我们造出了一个永久运动。同样,因为这个机器,是可反转的,所以,通过让这个机器反向运行,我们也可以证明,1英尺不能超过3X。所以,3X既不比1英尺大,也不比它小。所以,通过这个论证,我们就发现了规律,即X=1/3英尺。可以清晰地概括如下:运行一台可反转的机器,让一磅重量降下一定距离,那么,这个机器,就可以把p磅的重物,抬高此一定距离的1/p。表示这一结果的另一方法就是,三磅乘以抬高的高度,在我们这个问题中就是X,等于一磅乘以降下的距离,这里就是1英尺。如果我们让所有的重量,乘以它们现在的在地板之上的高度,然后,让机器运行,再然后,让所有的重量,再乘以所有的高度,将没有任何变化。(在我们的例子中,我们是把一个重量降低,抬高了几个重量,对此例子,我们应该归纳一下,即我们实际只移动了一个重量,然而,这种归纳,并不容易。)
[A1]