hello,
how can I calculate the moment of inertia of a crankshaft-connecting rod system ?
I've been told to assume 1/3 of connecting rod mass to roughly compute its
rotational moment of inertia. Are there formula to perform precise computation ?
I searched both Usenet and the Web but found nothing useful.
Thanks
Google, Advanced
all the words: crankshaft connecting rod formula
the exact phrase: moment of inertia
113 hits. A reasonable number, that a few minutes of reasearch can wade
through. Your English seems to be good enough...
David A. Smith
total or at any instant?
you need to sum more than one rod and ofset only if your time reference of
your purpose is longer than the cycle of one rotation - or for curiosity
- the flywheel is sized on other parameters, and anyway, it too only uses
one rod and offset as its prime driver.
do one rod and one offset - for total, times the number of pistons rotating
within the time frame - and don't forget
1) the rotational mass moment of inertia since the rod rotates on the pin and
shaft
2) the inertia of the crank offset
3) the connecting pin
for rotational, use the pin as the reference point
for linear, use the block.
for mass moment, only rotational
rotational moment of inertia
Odd - since the mass moment is a cube function of length and straight-up on
mass, and since the calculation is a fairly simple calculus problem.
Why would a professor assign a system set of inertials, which is fairly
complex, and then set an erroneus assumption on a simple calculation?
Tell me it's not another professor who lacks real world experience..
no, he's not a professor, he's an engineer :-)
To further clarify the issue, I have to compute rotational moment of
inertia about
shaft axis (one cycle and one piston) of crankshaft+con rod system.
Crankshaft alone is not a problem, the problem is the connecting rod,
because
its movement is a complex one, neither rotational nor linear.
How much its rotational component affects crankshaft rotational moment
of inertia,
so what is moment of inertia of the whole crankshaft+con rod system
about shaft axis ?
The advice I got to simplify computation is to assume 1/3 con rod mass
as it was in the
pin, rotating about shaft axis, and 2/3 as it had linear alternate
movement.
So I computed moment of inertia about shaft axis of the
crankshaft+assumed mass in the pin.
I wonder if there is anything more precise than such rough
computation.
Thanks
Maurizio
snipped-for-privacy@aol.comnono (Hobdbcgv) wrote in message
I think you mean BOTH rotational and linear
All complex problems use the same approach in their solution - one part at a
time, one reference point,dditive or multiplicative interaction
Thus the connecting rod is a mass acting at a distance from the axis of the
crank.
It is also a mass-moment multiplied by a lever arm which in turn acts thru an
inefficient connection.
One step at a time. No easy way.
.
I believe the missing concept here is "reflected
inertia" -- the apparent rotational inertia due to loads
being driven by a shaft thru other mechanisms--a gear
reduction or lead screw, for example. Or the slider-crank
in this case, which is a much uglier problem if an exact
solution is required.
Ned Simmons
I am familiar with the phenomena, although "reflected inertia" is not the
term we used.
(I once built a machine to put reflected impedance, as we called it, back
into the toytoa IC engines.
And I have had experience with, and designed for, the transfer of electric
motor's mass moment "backwards" thru gear sets on cranes, booms, and hoists.)
Thinking out loud, for discussion - and don't bet the house on this, this
is the internet and you get what you pay for -
I do not believe that reflected inertia is the condition here, although
with the piston driving the system rather than the crank driving the system,
moving the reference point to the wrist pin would be more accurate than using
the crank center and would minime calculations.
In other words, the moving piston sees the inertia of the crank, flywheel
and drive train at the pin; the crank "sees" no inertia, like the rod, it is
part of it.
In this approach, the torsional frequency of the crank and flywheel system
is then used to compare the force along the stroke with the torsional cycle to
determine if crank torsional deflection and attendant energy storage and
release adds any "reflected impedance" to the peak impedance of the force
chain seen by the piston.
In lay terms, the combustion force pushes on the rod and winds up the crank
and the crank unwinds at its natural frequency while input force is lower yet
still inputting, potentially giving more force on the crank and rod than
combustion alone gives. This is usually minimized by mismatching the frequency
of the crank and the piston travel
With the disconitnuity when the rod shifts across the crank-rod bearing gap
during the rotation, you can get jerk - But for the inertia problem, the
discontinuity generally becomes moot because when the rod is not connected,
it's inertia felt by the crank is nil.
imho
On 6 Sep 2004 08:50:54 -0700, mauri snipped-for-privacy@libero.it (Maurizio) wrote:
Sum the contribution of each particle of the crank and con rod times
its distance from the axis of rotation for each increment of rotation
between 0 - 90 degrees.
Brian Whatcott Altus OK
The rods will have different moments of inertia about different axes,
normally you might be interested in it's moment about the little end, big
end, or it's centre of gravity.
To get any of these, break down the design of the rod into primitive
elements, ( cylinders, cubes,... ) use their inertia's ( from your reference
book ) and the parallel axis theorem to add them together for your rod.
Two rods of the same weight can have different inertia's ( one could have
heavy ends and a light beam compared with an other ).... so you need details
of the rods designee.
Another factor which affects the importance of inertia in an analysis is the
angle through which a rod swings, which depends on the stoke and the rod
length, a long rod will swing less, so the effects of inertia will be
smaller in relation to it's mass.
If you have been given an "out " from these complications by being given an
" add 1/3 to the mass of the rod and forget inertia " heave a sigh of relief
and take it :-).
For a detailed analysis of the system I think you also need the position of
the rod's centre of gravity.
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