事实上,所谓“逐差法”不过是放大了的“邻差法”(△x=aT^2),对x1,x2,...x6;不过是将前三段看作一个X1=x1+x2+x3,后三段看作一个X2=x4+x5+x6,这两段的时间为3T,根据邻差法,
a=(X2-X1)/(3T)^2=(x6+x5+x4-x3-x2-x1)/9T^2
=[(x6-x3)+(x5-x2)+(x4-x1)]/9T^2==>
这与3a1*T^2=(x4-x1);3a2*T^2=(x5-x2);3a3*T^2=(x6-x3)==>
a=(a1+a2+a3)/3=(1/3)*{[(x4-x1)/3T^2]+[(x5-x2)/3T^2]+[(x6-x3)/3T^2]}=[(x6-x3)+(x5-x2)+(x4-x1)]/9T^2推出的结果是一样的.
而按照你的上述理解,推出的结果就不对了.