2-E Conservation of momentum
 

2nd para "The proportion of the separation rate to the approach rate expresses the elasticity involved, and is called the Coefficient of Restitution which is 80% for better golf balls - but drops below 70% at high speeds..."

3rd para "This means that even with precise geometrical alignments the ball will not separate from the club at more than 80% of their approach speed. That produces, roughly, this condition - the Clubhead approaching Impact at 100 MPH has slowed to 80 MPH at separation."

Ok so far

"The ball leaves the club at about 70 MPH (70% of 100 MPH approach speed). "

Ok -- if 100 mph is considered a 'high speed' as above in 2nd para, but why mention it again!

"To do this the ball must be travelling 150 MPH."

150 mph ?? - I don't see the logic!!
PLEASE EXPLAIN

"If the Clubhead speed at separation is 40 MPH, the ball can only travel 110 MPH under this law."

Agreed if 40 mph is 70% of 110 mph, sounds about right, but why repeat it?

"Notice this - the ball acquires only 70% of the Clubhead "approach" speed (so there must be some speed) but 100% of the Clubhead separation speed (so there must be resistance to
deceleration)."

It doesn't "acquire" separation speed - the speed at separation is less than the approach speed because of the Impact Interval.

3rd para "The divergent Impact and Separation Vectors (2-C-1/2/3) are always equal in..."

I can find no such vectors in those drawings nor in the 'Vector Symbol Legend'. Where are they?

Thanks in advance,
Alan (22-10-05)

The ball gets 70% of clubhead impact velo, which is 70 mph if the impact velo is 100 mph. In addition it gets 100% of clubhead separation velo, which is at best 80% of impact velo due to energy losses. So assuming the best case, this gives a separation velo of 80 mph. So ball speed is 70 + 80 = 150 mph.

Thanks indeed, now I understand.