A scale can only yield an answer whether the left or right sides are heavier. I would use the divide and conquer binary strategy. My answer would be 3 attempts. First attempt: Put four on the left and four on the right. Which ever side weight the most I would keep these 4 balls and discard the other 4 balls. Now I would take group of 4 heavier balls and divide them into two groups of two for the second weighing. Now I would get a second result of which group was heaviest. Then on the third attempt I would have on two balls so I would place one on the left and one on the right. Which ever one tipped the scale would the heaviest ball of the starting 8.

A scale can only yield an answer whether the left or right sides are heavier. I would use the divide and conquer binary strategy. My answer would be 3 attempts. First attempt: Put four on the left and four on the right. Which ever side weight the most I would keep these 4 balls and discard the other 4 balls. Now I would take group of 4 heavier balls and divide them into two groups of two for the second weighing. Now I would get a second result of which group was heaviest. Then on the third attempt I would have on two balls so I would place one on the left and one on the right. Which ever one tipped the scale would the heaviest ball of the starting 8.