To provide some perspective on the dimensions of atomic defects, consider a metal specimen with a dislocation density of 105 mm2 . Suppose that all the dislocations in 1000 mm3 (1 cm3 ) were somehow removed and linked end to end. How far (in miles) would this chain extend? Now supposethat the density is increased to 109 mm2 bycold working. What would be the chain length ofdislocations in 1000 mm3 of material?
ENGR 121 B Lecture Notes for 10/31/2016 Spencer Kociba ● Calculating Standard Deviation ○ u=mean, N=number of elements in a vector N ○ SD= 1 * ∑(x(i) − u) √ N−1 k=1 ■ Take each element, subtract the mean and square it. Repeat this for all the elements in the vector and sum them. Divide by N-1 and take the square root ○ Function this_std=standard_deviation(vec) ○ Part 1 (calculating the mean) ■ sum=0 ■ For i=1:length(vec) ● sum=sum+vec(i) ■ End ■ this_mean=sum/length(vec) ○ Part 2 (calculation of SD) ■ For i=1:len