Inertia maths help

On Jan 29, 8:26*pm, Chalo wrote:
Lou Holtman wrote:

Just go ride outside, saves you a lot of worries. ;-)

Lou, bright sunshine at 1 degree Celcius today.

In Austin Texas? *Mostly sunny and 21C at the moment.

So there.

Sounds awful! What's there to look forward to? ;-)

Joseph

On 1/29/2011 1:29 PM, Joseph Santaniello wrote:
On Jan 29, 8:26 pm, wrote:
Lou Holtman wrote:

Just go ride outside, saves you a lot of worries. ;-)

Lou, bright sunshine at 1 degree Celcius today.

In Austin Texas? Mostly sunny and 21C at the moment.

So there.

Sounds awful! What's there to look forward to? ;-)

Unbearable summer heat and humidity. Not at all like the pleasant
Norwegian summer.

--
Tºm Shermªn - 42.435731,-83.985007
I am a vehicular cyclist.

Joseph S wrote:

Chalo wrote:

Lou Holtman wrote:

Just go ride outside, saves you a lot of worries. ;-)

Lou, bright sunshine at 1 degree Celcius today.

In Austin Texas? *Mostly sunny and 21C at the moment.

So there.

Sounds awful! What's there to look forward to? ;-)

Mercilessly sunny and 40C, of course.

Chalo

*Not at all like the pleasant
Norwegian summer.

I wouldn't know. I was sick that day...

Joseph

On Jan 29, 8:36*pm, Chalo wrote:
Joseph S wrote:

Chalo wrote:

Lou Holtman wrote:

Just go ride outside, saves you a lot of worries. ;-)

Lou, bright sunshine at 1 degree Celcius today.

In Austin Texas? *Mostly sunny and 21C at the moment.

So there.

Sounds awful! What's there to look forward to? ;-)

Mercilessly sunny and 40C, of course.


Actually that has a certain appeal!

Joseph

On Jan 29, 10:03*am, Joseph S wrote:
On Jan 29, 3:39*pm, Dave Lehnen wrote:



Joseph S wrote:

Well, I'm at least amusing myself here... radius of the roller, not
diameter...

Joseph

I'm probably at fault for the confusion, since I mentioned the ratio of
diameters in the text, but then used the ratio of radii in the
calculation. The ratio is the same, but radius is easier to use in other
parts of the calculation, which is why I used that.

I don't have time to check your calculations before my ride this
morning, but I'll do a calculation for my CycleOps trainer and see what
I come up with later today.

Dave Lehnen

Have a nice ride!

I can't get the numbers to make any sense. Using the roller radius
(instead of diameter), I end up with a desired total inertia of .
000586 which is even less.

.335m wheel radius
.0275m roller radius
115kg effective mass

Joseph

Well, it turns out I screwed up the formula. Sorry for not checking my
work more thoroughly. While it passed the first sanity check of having
correct units, it didn't pass the second intuitive sanity check that
the wheel radius shouldn't have appeared on the right side of the
equation (the speed ratio matters, but terms from the wheel radius
cancel each other out). The revised, and I'm fairly certain correct
equation is:

m(b+r) + m(fwe) = [J(trn) + J(fly)] / r(trn)^2

With your numbers, the total trainer moment of inertia should then be
about 0.087 kg-m^2. One of many solutions to get this with a single
steel disc would be one of 0.1085 m radius (4.272") and 0.0508 m
thickness (2"). I think this is too big to fit your trainer.
Unfortunately, most common engineering materials aren't much heavier
than steel, brass is only a little heavier. I don't think you'd want
to pay for one made of tungsten, for instance. Possibly you could pour
molten lead in a hollow steel one (don't breathe the vapors).

Sorry for wasting your time with the original incorrect equation.

Dave Lehnen

On Jan 29, 10:03*am, Joseph S wrote:
On Jan 29, 3:39*pm, Dave Lehnen wrote:



Joseph S wrote:

Well, I'm at least amusing myself here... radius of the roller, not
diameter...

Joseph

I'm probably at fault for the confusion, since I mentioned the ratio of
diameters in the text, but then used the ratio of radii in the
calculation. The ratio is the same, but radius is easier to use in other
parts of the calculation, which is why I used that.

I don't have time to check your calculations before my ride this
morning, but I'll do a calculation for my CycleOps trainer and see what
I come up with later today.

Dave Lehnen

Have a nice ride!

I can't get the numbers to make any sense. Using the roller radius
(instead of diameter), I end up with a desired total inertia of .
000586 which is even less.

.335m wheel radius
.0275m roller radius
115kg effective mass

Joseph


Well, it turns out I screwed up the formula. Sorry for not checking my
work more thoroughly. While it passed the first sanity check of having
correct units, it didn't pass the second intuitive sanity check that
the wheel radius shouldn't have appeared on the right side of the
equation (the speed ratio matters, but terms from the wheel radius
cancel each other out). The revised, and I'm fairly certain correct
equation is:

m(b+r) + m(fwe) = [J(trn) + J(fly)] / r(trn)^2

With your numbers, the total trainer moment of inertia should then be
about 0.087 kg-m^2. One of many solutions to get this with a single
steel disc would be one of 0.1085 m radius (4.272") and 0.0508 m
thickness (2"). I think this is too big to fit your trainer.
Unfortunately, most common engineering materials aren't much heavier
than steel, brass is only a little heavier. I don't think you'd want
to pay for one made of tungsten, for instance. Possibly you could pour
molten lead in a hollow steel one (don't breathe the vapors).

Sorry for wasting your time with the original incorrect equation.

Dave Lehnen

On 1/29/2011 4:53 PM, wrote:
On Jan 29, 10:03 am, Joseph wrote:
On Jan 29, 3:39 pm, Dave wrote:



Joseph S wrote:

Well, I'm at least amusing myself here... radius of the roller, not
diameter...

Joseph

I'm probably at fault for the confusion, since I mentioned the ratio of
diameters in the text, but then used the ratio of radii in the
calculation. The ratio is the same, but radius is easier to use in other
parts of the calculation, which is why I used that.

I don't have time to check your calculations before my ride this
morning, but I'll do a calculation for my CycleOps trainer and see what
I come up with later today.

Dave Lehnen

Have a nice ride!

I can't get the numbers to make any sense. Using the roller radius
(instead of diameter), I end up with a desired total inertia of .
000586 which is even less.

.335m wheel radius
.0275m roller radius
115kg effective mass

Joseph

Well, it turns out I screwed up the formula. Sorry for not checking my
work more thoroughly. While it passed the first sanity check of having
correct units, it didn't pass the second intuitive sanity check that
the wheel radius shouldn't have appeared on the right side of the
equation (the speed ratio matters, but terms from the wheel radius
cancel each other out). The revised, and I'm fairly certain correct
equation is:

m(b+r) + m(fwe) = [J(trn) + J(fly)] / r(trn)^2

With your numbers, the total trainer moment of inertia should then be
about 0.087 kg-m^2. One of many solutions to get this with a single
steel disc would be one of 0.1085 m radius (4.272") and 0.0508 m
thickness (2"). I think this is too big to fit your trainer.
Unfortunately, most common engineering materials aren't much heavier
than steel, brass is only a little heavier. I don't think you'd want
to pay for one made of tungsten, for instance. Possibly you could pour
molten lead in a hollow steel one (don't breathe the vapors).

Sorry for wasting your time with the original incorrect equation.

Dave Lehnen

Iridium is very dense and corrosion resistant.

--
Tºm Shermªn - 42.435731,-83.985007
I am a vehicular cyclist.

On Jan 29, 11:54*pm, " wrote:
On Jan 29, 10:03*am, Joseph S wrote:





On Jan 29, 3:39*pm, Dave Lehnen wrote:

Joseph S wrote:

Well, I'm at least amusing myself here... radius of the roller, not
diameter...

Joseph

I'm probably at fault for the confusion, since I mentioned the ratio of
diameters in the text, but then used the ratio of radii in the
calculation. The ratio is the same, but radius is easier to use in other
parts of the calculation, which is why I used that.

I don't have time to check your calculations before my ride this
morning, but I'll do a calculation for my CycleOps trainer and see what
I come up with later today.

Dave Lehnen

Have a nice ride!

I can't get the numbers to make any sense. Using the roller radius
(instead of diameter), I end up with a desired total inertia of .
000586 which is even less.

.335m wheel radius
.0275m roller radius
115kg effective mass

Joseph

Well, it turns out I screwed up the formula. Sorry for not checking my
work more thoroughly. While it passed the first sanity check of having
correct units, it didn't pass the second intuitive sanity check that
the wheel radius shouldn't have appeared on the right side of the
equation (the speed ratio matters, but terms from the wheel radius
cancel each other out). The revised, and I'm fairly certain correct
equation is:

m(b+r) + m(fwe) = [J(trn) + J(fly)] / r(trn)^2

With your numbers, the total trainer moment of inertia should then be
about 0.087 kg-m^2. One of many solutions to get this with a single
steel disc would be one of 0.1085 m radius (4.272") and 0.0508 m
thickness (2"). I think this is too big to fit your trainer.
Unfortunately, most common engineering materials aren't much heavier
than steel, brass is only a little heavier. I don't think you'd want
to pay for one made of tungsten, for instance. Possibly you could pour
molten lead in a hollow steel one (don't breathe the vapors).

Sorry for wasting your time with the original incorrect equation.

Dave Lehnen

Wasting my time?! On the contrary! Learned quite a bit and amused my
self to no end. To convey just how much fun, I considered what size
styrofoam flywheel I'd need...

One way I was thinking about it was to consider the KE of me at 9 m/s
and then when I sprint to 18 m/s it takes 10 seconds. At the end I
have more KE. Some of the energy I poured in goes to wind resistance,
the rest to acceleration. The fluid unit takes care of the wind
resistance, so all I need to do is absorb the KE difference. So I
played around with different FW dimensions as the matching RPM he
http://www.calculatoredge.com/mech/flywheel.htm trying to find
something suitable that also isn't crazy at 5 m/s worth of RPM. This
got me on the track of eliminating the wheel radius from the equation,
but I didn't go all the way.

I also thought about using a car or truck brake disc mounted to the
existing FW if I could find one of suitable dimensions. But do I
really want one of those spinning at 6000 RPM behind my leg? I figured
I'd make a shield of 2x4's.

Another idea whas a FW made of thin plates of the same dimension, and
riders of different mass could fit additional plates as necessary.

Thanks for the help!

Joseph

--
Tºm Shermªn - 42.435731,-83.985007

I mistook the hyphen for an unary operator and thought to myself, Good
God, where has Tom moved to now?!?