Rob E
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Registered: 1st Jan 06
Location: Madeley, Stafford....I want to live back in Wales!
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Its not as simple as removing 10kg rotational mass is the equivilent of 20kg static mass. I made this assumption in my first uni presentation and one of my lectures asked me how I had come to that conclusion. Of course, I looked like an idiot then because I couldnt prove something I had read on a forum. However, 6 months later in the 2nd presentation I had the answers.
I made a spreadsheet to calculate the difference of removing mass from different rotational parts compared to static parts which you are welcome to have a play around with if you want. U2U me an email address and I'll fire it over.
Its based on an mx5 layout but has the interia calcs for props, flywheel, wheels, tyres etc, but if you just fill it in to suit the skyline you might be able to pull some useful information out of it.
When I calculated it based on the figures for my mazda, every 1kg of rotational mass removed was the equivalent to 12.8kg static mass
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Ian
Site Administrator
Registered: 28th Aug 99
Location: Liverpool
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12 to 1?
Thats larger than 1.2/1.5 from page 1.
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AK
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Registered: 5th Jul 00
Location: Aberdeen City
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1 to 12 isnt right! way way off. Rotational mass of what?
I'd still like a copy though, ak@trackscotland.co.uk
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Ian
Site Administrator
Registered: 28th Aug 99
Location: Liverpool
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Bet you wish you'd not started with a barge car 
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AK
Member
Registered: 5th Jul 00
Location: Aberdeen City
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its lost over 200kg
But there are gains from using a GTR over an Impreza.
My Impreza is sOOOOoo much lighter though
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Rob E
Member
Registered: 1st Jan 06
Location: Madeley, Stafford....I want to live back in Wales!
User status: Offline
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quote:
Originally posted by Ian
12 to 1?
Thats larger than 1.2/1.5 from page 1.
Yes, thats effective mass removal which is based on effective inertia of an mx5 fitted with heaviest OEM stock wheels and then compared to the lightest available OEM stock wheel. If you were to play around with the spreadsheet and compare the mass of two prop shafts then I predict you will see a lot smaller static/rotational mass equivalent due to the fact the inertia of a prop shaft is massively smaller to that of a road wheel.
This was based on a series of calculations specific to a naturally aspirated mx5 which incorporates your gear ratios, final drive and the tractive effort of the car and then calculating the rate of acceleration.
Based on my calculations (and assuming an instantaneous gear shift) when you removed the equivilent mass from the wheels (rotational) and then compared it to removing it from say, the body, (static) the acceleration times improved by 0.09 seconds. It may be small but if you remove the mass from the rotational side of the vehicle you will gain a larger advantage than removing the mass from a static side.
If anyone else want a copy of the spreadsheet then I dont mind sending it. Its quite interesting if stuff like this makes you tick 
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LeeM
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Registered: 26th Sep 05
Location: Liverpool
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i think it depends on diameter and speed, the higer each are the more of an effect the rotation has on the mass. or something
[Edited on 05-01-2013 by LeeM]
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antnee
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Registered: 30th Dec 07
Location: Cov Drives: Clio 197
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I was going to comment saying there isn't a 'figure' that can be applied to all situations, many things effect it such as mass, mass distribution, RPMs, diameter, but Rob has pretty much said all of that.
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AK
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Registered: 5th Jul 00
Location: Aberdeen City
User status: Offline
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yup... no clear cut answer, didnt think there would be... but though an approximation would be possible.
Its a heavy thin shaft that spins fast... not a huge gain to be had but still weight is weight
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