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Author Topic: RG?  (Read 2588 times)

vadertyme03

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RG?
« on: April 08, 2009, 04:58:03 AM »
Would anyone happen to know in what units RG is in? Have a class that talks a little about radius of gyration and was curious.
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sspeckma

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Re: RG?
« Reply #1 on: April 08, 2009, 01:28:20 PM »
Gee let's see it's a radius maybe it has units of Length?
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nospareball

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Re: RG?
« Reply #2 on: April 08, 2009, 02:20:23 PM »
Inches
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Juggernaut

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Re: RG?
« Reply #3 on: April 08, 2009, 02:32:48 PM »
RG is a number that is indicative of the distance the mass rotates from the center of the balls rotation. Being a measurement of distance, I would assume it is units of length.

  An example would be a ball with a high RG value of 2.6 and a low RG value of 2.4 ( for demonstration purposes ) The 2.6 value would be achieved by having the top of the weightblock ( located by the "pin" ) in the balls initial track and the 2.4 value would be achieved by having the top of the weightblock ("pin") on your positive axis point.  The difference between the two values is known as the RG differential.


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J_w73

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Re: RG?
« Reply #4 on: April 08, 2009, 02:42:33 PM »
I understand it as ...

in my words..

say you have a 16lb ball that has a 2.6 rg.. that means that the force required to rotate this ball would be the same as if you had a ball that was 2.6 inches in radius that had all of its 16lbs of weight spread out about the surface..

in more technical terms

Technically speaking, the radius of gyration is defined as the square root of the moment of inertia divided by mass of the object. Therefore, the radius of gyration is the distance that, if the entire object's mass were together at only that specific radius, would yield the same moment of inertia. The moment of inertia for an object is the ratio of applied torque and the resultant angular acceleration of the object. Translating the physics definition, the moment of inertia measures how easy an object will rotate when a force is applied. In simple terms, the radius of gyration determines how easy it is for the bowling ball of particular weight to rotate about a given axis and is a measurement of where the weight is located inside the ball relative to the center.



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Edited on 4/8/2009 2:43 PM
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RyanRPS

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Re: RG?
« Reply #5 on: April 08, 2009, 07:51:30 PM »
quote:
its seconds in which the ball takes to rotate. the lower the rg the quicker it revs, the higher etc..

its based off of a machine, so they arent fixed numbers. which is why some companies show minimum and maximum RGs.

Edited on 4/8/2009 2:09 PM


This made me smile

Radius of Gyration is the square root of the moment of inertia about an axis devided by the mass of the object.. the reason that there is a maximum and a minimum is because you can work out the moment of inertia through any axis... so you can work out a maximum and minimum

(so basically, to think of max and min RG, if you have a football shaped core, you take one core lenth-ways and one width ways for max and min )

Ryan

Oh and sorry yes, its a unit of distance!



Edited on 4/8/2009 8:02 PM
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dizzyfugu

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Re: RG?
« Reply #6 on: April 09, 2009, 03:03:09 AM »
quote:
quote:
its seconds in which the ball takes to rotate. the lower the rg the quicker it revs, the higher etc..

its based off of a machine, so they arent fixed numbers. which is why some companies show minimum and maximum RGs.

Edited on 4/8/2009 2:09 PM


This made me smile


Me too!

As a different approach, image the RG numbers as distance of the ball's weight distribution from the center (through which the rotational axis goes). The lower the RG number, the more center-heavy the ball/core is, and the lower the moment of inertia becomes - the ball revs up more easily (both with the hand as well as on the lane upon friction).

Beyond that, Juggernaut has given a good explanation of the rest, what RG max, min and the RG diff. mean.

Additionally, the mass bias strength is calculated the same way, it is just the second "strongest" axis at 90° from the pin axis, and the mass bias stregth is the respective RG differential.
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leftyinsnellville

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Re: RG?
« Reply #7 on: April 09, 2009, 07:53:32 AM »
Does anyone know the lowest and highest rg balls ever made that were approved?

How about the ball with the highest differential ever that was approved?

I'm sure they wouldn't be legal today, but I was just curious.
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dizzyfugu

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Re: RG?
« Reply #8 on: April 09, 2009, 08:48:46 AM »
If I remember correctly, legal RG min (still) is 2.45", RG max at 2.65", max diff. used to be 0.08" and has now been reduced for any newly released ball to 0.06" now. Visioanry's Immortals were the last balls to exploit the diff. max, the SD-73 was another flare monster, and I have seen older overseas balls with differentials close to the 0.08" limit when it was still legal.
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vadertyme03

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Re: RG?
« Reply #9 on: April 09, 2009, 10:10:19 AM »
I understand it's in units of length. Was wondering more specifically what units of length. I see someone said inches. Does anyone else know this to be true?
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J_w73

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Re: RG?
« Reply #10 on: April 09, 2009, 10:31:35 AM »
quote:
I understand it's in units of length. Was wondering more specifically what units of length. I see someone said inches. Does anyone else know this to be true?
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When refering to bowling balls that are described in the USA it is inches.
I say this because other countries may convert to a metric measurement...I don't know..
but all the numbers that are given and talked about here are in inches.
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RyanRPS

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Re: RG?
« Reply #11 on: April 09, 2009, 10:57:43 AM »
yes, it is in inches when written like "2.52"...

its pretty simple to work out.. take for example if you wanted to work out the maximum possible RG for a solid plastic ball weighing 15lbs (not a "polyester spareball", a solid plastic for experimental reasons)....

radius = 4.297 inches
mass = 15 lbs

I=2/5.m.r.r = 110.795

RG=SQRT I/M = 2.718

therefore for any ball made of any one material (mass gets cancled out in the equation) and measures that of a bowling ball, the maximum possible RG of 2.718 (inches)

Ryan
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vadertyme03

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Re: RG?
« Reply #12 on: April 09, 2009, 12:23:44 PM »
Thanks a lot for all the replies.
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