Omega Owners Forum
Chat Area => General Discussion Area => Topic started by: amazonian on 20 January 2018, 15:39:35
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Can any of you guys answer this?
A cylinder or disc (say the rough shape of a small roll of cellotape) has a diameter of 500mm.
It is 51mm long and it has a bore of 203mm.
The question is, if the diameter of 500mm was reduced by 18mm, what percentage of mass would be lost?
I have tried twice and been wrong both times, and cant think what I am missing (apart from numerous brain cells)
:) ;)
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AreaT equals 2(3.142x250)-2(3.142x101.5)
Volume equals AreaT x 51.
Are you losing 18mm from the outside diameter or is the bore increasing by 18mm?
Your answer will depend on which... ;)
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Yes, losing 18mm from the outside diameter.
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OK, new equation is:
AreaT2 equals 2(3.142x241) - 2(3.142x101.5)
New VolumeT2 equals AreaT 2 x 51.
The difference is obviously VolumeT - VolumeT2 :y
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Don't forget, the area of a circle is π r².
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Rookie mistake :-[
2piR is the circumference ::)
Adjust the math accordingly, but the order of the workings is sound ;)
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Rookie mistake :-[
2piR is the circumference ::)
Adjust the math accordingly, but the order of the workings is sound ;)
Maths , were English not American old chap :y
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Rookie mistake :-[
2piR is the circumference ::)
Adjust the math accordingly, but the order of the workings is sound ;)
Maths , were English not American old chap :y
Who was?
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Ha ha! Touché .
:) :)
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AreaT equals 3.142(250x250)-3.142(101.5x101.5)
VolumeT equals AreaT x 51.
AreaT2 equals 3.142(241x241) - 3.142(101.5x101.5)
New VolumeT2 equals AreaT 2 x 51.
The difference is obviously VolumeT - VolumeT2 :y
In answer to the original question... ;)
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Ha ha! Touché .
:) :)
Leave the door open wide enough and someone will eventually walk through it :P
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Dr Gollum
It is so long since my school maths lessons I cannot even remember what to do with figures in brackets etc.
One of my grandsons lives in Finland and his maths book shows a completely different method to my English grandsons, and both are different to my own vague memories.
I made the answer 6.1%, my son says it is 8.4%. I don't know what you made it for the above stated reason, but thank you very much for taking the trouble to reply.
:y :)
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Can any of you guys answer this?
A cylinder or disc (say the rough shape of a small roll of cellotape) has a diameter of 500mm.
It is 51mm long and it has a bore of 203mm.
The question is, if the diameter of 500mm was reduced by 18mm, what percentage of mass would be lost?
I have tried twice and been wrong both times, and cant think what I am missing (apart from numerous brain cells)
:) ;)
Ignore the length - it is common to both cases.
Using area = pi.r2
case 1 = pi x 2502 - pi x 101.52
case 2 = pi x 2412 - pi x 101.52
percentage loss = [case 1 - case 2] / case 1 x 100%
I make it 8.46% (rounded to 2 decimal places)
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A roll of cellotape has a plastic & glue outer and a cardboard centre. Reducing the diameter by removing some/all of the tape won't alter the mass in the way these sums predict due to the differences in density of the two materials.
So, is the disk a constant density? ::)