Spinner Energy

[stinger]

Elsewhere on the FRA site, you may have seen that Im working with James Baker on some guidelines for Spinning Weapons. We have a neat way of assessing the Kinetic Energy in your designs which shows how fast we would let you run. We only need to make a crude assessment of Style (mass distribution) + measured figs for actual Mass & Dia.

Now, at present the FRA are NOT sanctioning use of spinners except under controlled conditions in Class 2 arenas. James and I want to set sensible rules that will allow more spinners, albiet with a limit on permissible rpm.

We suggest that you start designing them with alternative drive systems - one geared down for live events, and another geared up for serious combat. You never know, the reduced spin-up time with a low ratio might even be an advantage ! Please note, that we do not anticipate permitting electronic speed limiters (but would allow reduced voltage).

We need to collect as much data as possible to decide where we should set the level for different arenas. For example we believe the most aggressive spinner is up at around 38KJ, with the most well know in the region of 16KJ, and Feathers peaking at 5-6KJ.

We have some experience of 6KJ in Jonnos arena, and want to collect as much additional info as poss to help establish challanging, but manageable poower limits.

Please post details here, or if you dont want to go public, then send them direct to me or James.

We would like to see Basic Dimensions of the spinning element plus its Mass and anticipated RPM. If you have a calculation for KE - then please post that as well.

Thanks,
Kevin S

[storm]

Spinner Energy

Vortex - 3Kg
4500rpm
Diameter 200mm - almost all mass at the edge of rim - it only has 2 spokes on the disk.

No idea what energy that translates into, but hope the figures are of some use.

Ed
http://www.stormrobot.comhttp://www.stormrobot.com

[robotwars]

Spinner Energy

Strip - Feather weight - 3.5Kg disc
3500rpm
Diameter 500mm - mass equally distributed apart from two 0.25kg teeth on outer edge.

Jonno
http://www.roamingrobots.co.ukwww.roamingrobots.co.uk
http://www.ukinnovations.co.ukwww.ukinnovations.co.uk

[ewan]

Spinner Energy

Big Daddy- Featherweight- (revised disc from my profile pic) 2kg Bar at 3000rpm, 500mm dia, vertical
with 0.8kg mallets at the end of each side.

[revolutionroger]

Spinner Energy

Think I can be of a little help here;
Vortex generates 3KJ of kinetic energy
Strip produces 8KJ of kinetic energy (@3500rpm)

Rough figures based on info above

[big_nipper]

Spinner Energy

Looking at Strip and votex I think those figures sound wrong but what to I know. I dont know how to work it out either, but I do know that 90% of the mass of vortexs disc is on the rim and strips is evenly distrbuted. Going off the size of the discs i would say if the speeds are true they are very even in power. I wouldnt say that its 2.5 times the power of vortex.

Prove me wrong.

[revolutionroger]

Spinner Energy

Does look wrong doesnt it? But it isnt (based on the figures above (assuming Jonnos can actually do 3,500rpm)

It has to be remembered that KE is proportional to the SQUARE of velocity. (KE=1/2MV^2)

Unlike Momentum which is directly proportional to velocity (P=MV)

Since Kinetic Energy is proportional to the Square of the Velocity, this means that a Mass traveling at 2 metres per second has 4 times the Kinetic Energy of the Same Mass travelling at 1 Metre per second, Even though it only has double the momentum.

Perhaps measurements should be compared in terms of momentum (Kgm/s) and not Kinetic Energy (KJ)?

Anyone please feel free to correct me on any of this, I am by no means skilled in this field.

Roger.

[geoff]

Spinner Energy

Roger,
Yes you are right in princable but we are dealing with rotating mass not mass traveling in a straight line!
There are questions involving moments of inertia to contend with as well as angular velocity.
I also confess to be a bit rusty but as I remember it the kinetic energy of a rotating mass is somthing like..


Kinetic energy = ½ moment of inertia X angular velocity squared

or E = 1/2Iw^2


I hope this helps, although as I said its been a long time since I really had a use for maths like that so dont take it as gospel!

Geoff,
Team Scorpion.

[revolutionroger]

Spinner Energy

Sorry, yes KE=1/2MV^2 is for mass in a straight line - didnt mean for anyone to use it as it will not calculate KE of rotating mass - but does clearly show how KE differs from Momentum which is how most people think.

[geoff]

Spinner Energy

Been giving this a little more thought, The calculation of angular velocity is relatively simple. But the problem I can see in trying to calculate KE in a rotating weapon is going to be determining the moment of inertia. It will totally depend on the construction of the weapon . How mass is distributed through the disc will have a serious effect on the value. I think (and I may be wrong) but the only way to arrive at an accurate value will be to calculate it for every component of the disc and add it all together to arrive at a total value. Which is fine if you could weigh all the component parts separately, this would also mean having to calculate their velocity which of course is relitive to their radius from the center but how you calculate that once the disk is built, I have no idea!
Maybe some one who understands this better than I may have a solution. I would love to know what the answer is!

[jim]

Spinner Energy

This is just an idea off the top of my head: You could produce a maximum moment of inertia, based on the assumption that all the weight is on an infinitely thin band at the circumference of the disc. It wouldnt be perfect but it would be better than no measurement at all.

[revolutionroger]

Spinner Energy

The only other way that you could measure a disks stored energy would be to measure the work required to accelerate the disk to a fixed velocity over a fixed time. Havent thought this through yet, but once a formula is thought out and equipment set up, measuring a new disk would simply be like balancing a wheel.

Not saying its easier, but should be more accurate, and require only the initial effort as opposed to having to take each disk from scratch, and as you say calculating each change of shape/part as you would have to. This may be easier for one or two, but over time still being laborious.

Dont have all the answers.
Roger

[s8nsmum]

Spinner Energy

MOI of a disc rotating about its centre point = 1/2 x Mr^2
MOI of a bar rotating about one end = 1/3 Ml^2
where
M = Total Mass
r = radius
l = length

If the mass is not rotating about its centre of gravity then you have to use the parallel axis theorem to calculate the additional MOI. This is
(Total MOI) = (MOI about centre of gravity) + (Area of object) x (distance of axle from centre of gravity).

so for vortexs disc if modeled as a large disc (200mm) with a smaller disc cut out of the middle (140mm?) and a bar put across (width 30mm) then total x-sectional area = 0.068 m^2
mass per unit area = 43.9 kg/m^2

Total MOI = 0.03 kgm^2

Total KE = 3.3 KJ

Similarly for strip:

MOI = 0.095 (seems a bit large)

KE = 6.4 KJ

Ive just done these in a rush and am pretty sure one of them is wrong .If Anyone wants me to explain my methods or any thing then just post your e-mail or e-mail me and i will send you the full calculations

Joe Townsend

[s8nsmum]

Spinner Energy

Also other ways to calculate moment of inertia are
1: to take the disk off, place it on a rotational pendulum and calculate the difference in period.
2: place the disc and axle on two bars at a slight decline. By timing the disc rolling measured distance at a measured angle you can calculate the MOI. If anyone wants a full explanation of these method please feel free to e-mail me and I will dig out my 2nd year lab books.

Joe Townsend

[geoff]

Spinner Energy

I found this,
http://homepages.which.net/~paul.hills/Spinningdisks/Disks.htmlhttp://homepages.which.net/~paul.hil...sks/Disks.html
I think that covers it all!