How is true zero imbalance achieved/determined when using a soft bearing driveshaft balancer?
To my understanding, in a perfect situation, both the head stock and tail stock spindles spin at .000" imbalance with a yoke attatched. When a driveshaft is bolted in, and is spinning at .005" imbalance, weight is added and moved around until the dial then reads .000" imbalance. The driveshaft is now perfectly balanced.
In a non perfect situation, because actually getting the dial to read .000" imbalance with just a yoke spinning on the spindle is very time consuming, say it reads .001" before a driveshaft is bolted in. After the shaft is bolted in, and it reads .005" imbalance, is the net total imbalance technically only .004"? Does weight added to the shaft only have to bring the imbalance down to the .001" reading that was there with only the yoke spinning?
Does making the shaft spin at .000" when the yoke alone spins at .001" leave the shaft unbalanced?
To my understanding, in a perfect situation, both the head stock and tail stock spindles spin at .000" imbalance with a yoke attatched. When a driveshaft is bolted in, and is spinning at .005" imbalance, weight is added and moved around until the dial then reads .000" imbalance. The driveshaft is now perfectly balanced.
In a non perfect situation, because actually getting the dial to read .000" imbalance with just a yoke spinning on the spindle is very time consuming, say it reads .001" before a driveshaft is bolted in. After the shaft is bolted in, and it reads .005" imbalance, is the net total imbalance technically only .004"? Does weight added to the shaft only have to bring the imbalance down to the .001" reading that was there with only the yoke spinning?
Does making the shaft spin at .000" when the yoke alone spins at .001" leave the shaft unbalanced?