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+--- Thread: Answered Left spin vs Right spin? (/Thread-Answered-Left-spin-vs-Right-spin)
Left spin vs Right spin? - HyperBey123 - Aug. 21, 2018
In burst Beyblades make each other skip teeth and then, burst.
But a left spin vs right spin match would make them skip teeth backwards due to the fact the bursting gimmick requires both beys pushing in the same direction to actually make them skip teeth. Despite this, I have seen left vs right spin battles where one of them bursts and some where they act according to my thinking.
RE: Left spin vs Right spin? - 1pratham - Aug. 21, 2018
Recoil.
a bit of physics, let the speed of slower spinning bey be X, and Speed difference be y
the faster spinning beys has speed, x + y , which is greater than by Slower spinning by y.
So if you Subtract Slower speed from both,
i.e., x-x=0; x+y-x= y
that means at some point of contact Slower spinning bey will act as stationary object, and faster bey will skip it's teeth out recoil from "stationary" bey
RE: Left spin vs Right spin? - g2_ - Aug. 21, 2018
(Aug. 21, 2018 3:40 PM)1pratham Wrote: Recoil.
a bit of physics, let the speed of slower spinning bey be X, and Speed difference be y
the faster spinning beys has speed, x + y , which is greater than by Slower spinning by x.
So if you Subtract Slower speed from both,
i.e., x-x=0; x+y-x= y
that means at some point of contact Slower spinning bey will act as stationary object, and faster bey will skip it's teeth out recoil from "stationary" bey
Bravo, great post.
RE: Left spin vs Right spin? - 1pratham - Aug. 21, 2018
(Aug. 21, 2018 3:45 PM)g2_ Wrote:(Aug. 21, 2018 3:40 PM)1pratham Wrote: Recoil.
a bit of physics, let the speed of slower spinning bey be X, and Speed difference be y
the faster spinning beys has speed, x + y , which is greater than by Slower spinning by x.
So if you Subtract Slower speed from both,
i.e., x-x=0; x+y-x= y
that means at some point of contact Slower spinning bey will act as stationary object, and faster bey will skip it's teeth out recoil from "stationary" bey
Bravo, great post.
Thanks ;D
Beys are full of physics
RE: Left spin vs Right spin? - GOD CHIPS - Aug. 21, 2018
(Aug. 21, 2018 3:40 PM)1pratham Wrote: Recoil.
a bit of physics, let the speed of slower spinning bey be X, and Speed difference be y
the faster spinning beys has speed, x + y , which is greater than by Slower spinning by y.
So if you Subtract Slower speed from both,
i.e., x-x=0; x+y-x= y
that means at some point of contact Slower spinning bey will act as stationary object, and faster bey will skip it's teeth out recoil from "stationary" bey
This thing applies when somebody is using Drain Fafnir ( rubber appilies recoil to other beys when DF is launched lightly )
RE: Left spin vs Right spin? - 1pratham - Aug. 22, 2018
(Aug. 21, 2018 5:57 PM)GOD CHIPS Wrote:(Aug. 21, 2018 3:40 PM)1pratham Wrote: Recoil.
a bit of physics, let the speed of slower spinning bey be X, and Speed difference be y
the faster spinning beys has speed, x + y , which is greater than by Slower spinning by y.
So if you Subtract Slower speed from both,
i.e., x-x=0; x+y-x= y
that means at some point of contact Slower spinning bey will act as stationary object, and faster bey will skip it's teeth out recoil from "stationary" bey
This thing applies when somebody is using Drain Fafnir ( rubber appilies recoil to other beys when DF is launched lightly )
Increases Friction with Soft Plastic.
Even the dragon Heads do the same
RE: Left spin vs Right spin? - Phoenix8Nt - Aug. 22, 2018
(Aug. 21, 2018 3:40 PM)1pratham Wrote: Recoil.
a bit of physics, let the speed of slower spinning bey be X, and Speed difference be y
the faster spinning beys has speed, x + y , which is greater than by Slower spinning by y.
So if you Subtract Slower speed from both,
i.e., x-x=0; x+y-x= y
that means at some point of contact Slower spinning bey will act as stationary object, and faster bey will skip it's teeth out recoil from "stationary" bey
Well i like ur thinking
Its great to so physics out here bcos i myself is bachelor in physics first year
So good eqn man
But not to be rude
If the bey is spun at an equal speed without any contact then it can be considered as
Same speed of spin - X
And the depth contact from opponent -Y
Then it would be said to be X =Y
But in case of rubber the momentum would be reduced to 0 by spin equalization
So anything cannot be calculated by theory
The external forces such as stadium,design of bey ,metal, would affect it ..
So anyways sorry if i sound nerd ish
Ur eqn is correct
Sorry if im being rude