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29 November 2024 01:50
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want to know how they work, or what they can be used for then you can leave your question here for others to answer.
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Question |
| Asked by: |
Shahriar Yazdani |
| Subject: |
Calculation of the tilting force in gyroscope |
| Question: |
how can we calculate the tilting force for a gyro with known inertia and R.P.M. ? |
| Date: |
8 July 2003
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Answers (Ordered by Date)
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| Answer: |
Robert Beal - 25/07/2003 20:17:49
| | | Let M=moment, I=moment of inertia of the flywheel about its axis of
spin, F=tilting force, m=mass of flywheel, r=radius of flywheel
R=length of the flywheel's axle (ie, distance to the flywheel's
pivot point), w=rate (angular velocity in radians per second) of
precession, y=spin (angular velocity in radians per second) of the
flywheel, g=acceleration due to gravity (ie, the gravitational constant, 9.8 meters per second).
Then:
M = Iwy
RF = Iwy
F = Iwy/R
for a solid flywheel, I = (1/2)mr*r
for a "flywheel" shaped like a ring I = mr*r
(where r*r = r squared)
Use kilograms, meters, seconds, and radians for units of
measure in these equations - and don't forget to convert RPM
to radians per second.
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| Answer: |
Robert Beal - 25/07/2003 20:18:29
| | | Let M=moment, I=moment of inertia of the flywheel about its axis of
spin, F=tilting force, m=mass of flywheel, r=radius of flywheel
R=length of the flywheel's axle (ie, distance to the flywheel's
pivot point), w=rate (angular velocity in radians per second) of
precession, y=spin (angular velocity in radians per second) of the
flywheel, g=acceleration due to gravity (ie, the gravitational constant, 9.8 meters per second).
Then:
M = Iwy
RF = Iwy
F = Iwy/R
for a solid flywheel, I = (1/2)mr*r
for a "flywheel" shaped like a ring I = mr*r
(where r*r = r squared)
Use kilograms, meters, seconds, and radians for units of
measure in these equations - and don't forget to convert RPM
to radians per second.
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| Answer: |
Robert Beal - 26/07/2003 03:17:48
| | | The above will give the force F in Newtons (using kilograms, meters,
seconds, radians, and g=9.8 meters-per-second in the calculations).
You could also do the calculations using units of pounds,feet,seconds
and radians-but be careful since pounds in this case is the unit of
force, so if mass is given in "pounds" you must divide by 32.2 to
convert the "mass given in pounds" to slugs. Also, g the acceleration
due to gravity (ie, the gravitational constant) using these units will
be 32.2 feet-per-second-squared (not 9.8) - then the final answer, the
force F will be in pounds.
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| Answer: |
vipin - 06/08/2003 03:42:55
| | | no
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| Answer: |
sh.yazdaqni - 07/08/2003 07:35:08
| | | Hi Robert Beal
thank you for your reply. but we do not have the w (the rate of precession). because the w depend on f(tilting force). i need a fomula independent of w for calculating tilting force.
thank you
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| Answer: |
Paul Tambala - 19/09/2003 04:09:51
| | | Hello.
First let me tell you that i do not have the equation you are asking for. You can back up your browser now or find out why i am bothering you with this answer...
Hi.....
I have had an interest in gyro for a while now and i have built some to understand their operation. I am a research physicist of 30 years experience and one thing i have learnt is that matematics is 'after the fact' when it comes to understanding the principals of a new phenomina. If you want to understand gyros, you first need a consistant model that makes sence, equations are really only required when you want to design a specific system that you already understand. Saying this, my model is almost complete and i have a brand new copy of mathamatica which i intend to aply to this issue in the near future.....
yours Tambala
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| Answer: |
Peter Boetzkes - 17/03/2004 08:33:56
| | | You indicated you want tilting force independent of precession. It is zero. If a gyro does not precess, it does not apply a tilting force. There may, however, be a tilting force due to gravity, or wind pressure, or some other externally applied force. The gyro will attempt to apply an equal and opposite tilting force by precessing at just the right rate.
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