golfbulldog
New
Hi There
It is My Understanding that Iron Byron has s Free Hinge which Attaches the Golf Club to Iron's Arm....... Suppose Iron Byron is Calibrated for a 100 mph Impact Velocity with the Club 90 Degrees to the Arm to Simulate Lag,,,,,, Now the Free Hinge is Locked in a Straight Arm Position to Simulate No Lag.... Iron is Fired with the Same Calibrations as before..... The Question is- What is the Impact Velocity in the No Lag Simulation?
Cheers
Billy - you must have been on enough forums to know that internet is pretty inefficient means of communication and is prone to misunderstanding.... 20 pages of stuff from a relatively straight forward question etc... so just to clear things up:
"same calibrations as before" - does that mean that the engine/rotor does exactly the same as it did to achieve the 100mph clubhead speed and you want to know the amount by which the the stiff arm swing is slowed ( by moving the mass away from the centre)... that is the way it read to me originally.
Now you say that the "Iron Byron arm speed is the Same for Both Cases"... but that would mean different settings ( at least as far as the engine were concerned)...no?
Now if you are saying that the motor applies whatever force is required to move the stiff arms at the same speed as the lagging swing ( programmed to achieve 100mph impact collision speed)... then you are asking an unusual question it seems.... the stiff armed swing has no release and there is no slowing down of the arms ( as long as the engine is set to do whatever it takes to achieve the speed you desire).... the lagging release does have a slowing down of arms and hands as the accumulator lag angle( left wrist cock) is released. So which section of the swing do you want the arm speed to be the same?
Sorry if I misunderstand your intention. What sort of answer are you expecting - a theoretical explaination / equations / a single numerical value???
If you can clear this up for us then cool.
Merci