Inertia maths help

Howdy All!

I'm deep in some winter training, and trying to do sprint workouts on
my new Kurt Kinetic Road Machine trainer. It's quite nice, but could
use a bit more inertia. There is an optional extra heavy flywheel
available, but they seem to be sold-out world wide, so I figure I
could get one turned out locally, and as a plus, I could have it's
mass tuned to suit me.

So I'm trying to figure out what flywheel weight and dimensions I
should spec. I'm most interested in sprint training, so I've been
trying to figure out my energy, but I'm not getting anywhere...

The Kurt does a good job of simulating wind resistance, so the trainer
speeds (and thus gearing) is similar to the real world, but the
trainer accelerates too fast.

Here is data for a typical (optimal) sprint I'd like to simulate:

Initial velocity: 35 km/h
Final velocity: 65 km/h
Duration: 10 sec
Distance: 150m
Total rider + bike mass: 115 kg

To complicate matters I do my sprints on a slight downhill that is
similar to the drop at a velodrome. About 5m.

I've guesstimated my acceleration (falsely assuming constant) to be
about .85 m/s/s ((18m/s - 9.7m/s) / 10s), and thus a force of about
98N. Given the 150m, that means 98*150= 14.7kJ. That makes sense. My
powertap tells me 13kJ but it doesn't know anything about the hill.

And that's where I get stuck. I can't figure out how to spec a
flywheel that can absorb that amount of energy at the appropriate
rate. Anybody feel like helping out?

Thanks,

Joseph

Joseph S wrote:
Howdy All!

I'm deep in some winter training, and trying to do sprint workouts on
my new Kurt Kinetic Road Machine trainer. It's quite nice, but could
use a bit more inertia. There is an optional extra heavy flywheel
available, but they seem to be sold-out world wide, so I figure I
could get one turned out locally, and as a plus, I could have it's
mass tuned to suit me.

So I'm trying to figure out what flywheel weight and dimensions I
should spec. I'm most interested in sprint training, so I've been
trying to figure out my energy, but I'm not getting anywhere...

The Kurt does a good job of simulating wind resistance, so the trainer
speeds (and thus gearing) is similar to the real world, but the
trainer accelerates too fast.

Here is data for a typical (optimal) sprint I'd like to simulate:

Initial velocity: 35 km/h
Final velocity: 65 km/h
Duration: 10 sec
Distance: 150m
Total rider + bike mass: 115 kg

To complicate matters I do my sprints on a slight downhill that is
similar to the drop at a velodrome. About 5m.

I've guesstimated my acceleration (falsely assuming constant) to be
about .85 m/s/s ((18m/s - 9.7m/s) / 10s), and thus a force of about
98N. Given the 150m, that means 98*150= 14.7kJ. That makes sense. My
powertap tells me 13kJ but it doesn't know anything about the hill.

And that's where I get stuck. I can't figure out how to spec a
flywheel that can absorb that amount of energy at the appropriate
rate. Anybody feel like helping out?

Thanks,

Joseph

I'm going to ignore the hill, at least at first, and look at matching
effective mass. A passive trainer will never simulate a hill very well
in any case.

With the bike on the trainer, the total effective mass comes from the
rear wheel and drivetrain, the rider's legs, and from the moment of
inertia of the rotating parts of the trainer, including its flywheel.
When on the road, the total effective mass does not include anything
from the trainer, but adds the mass of the bike and rider, and the
effective mass from the moment of inertia of the front wheel. Since the
contribution from the rear wheel, drivetrain, and rider's legs is
identical in both cases, you don't need to worry about these. (The
contribution from effective moment of inertia of the legs would be a
complex calculation, but not a very big part of the total anyway.)

So to match effective mass, the total moment of inertia of the trainer
should make up for the rider mass, and bike mass (complete with both
wheels), and effective mass from the moment of inertia of the front
wheel.

The trainer spins faster than the bike wheel, by the ratio of the rear
tire diameter to the friction roller diameter on the trainer. The
effective mass from the trainer varies with the square of this ratio.

let:
J(fw) = moment of inertia of front wheel
J(trn) = moment of inertia of trainer before new flywheel
J(fly) = moment of inertia of added flywheel
r(w) = radius of wheel, front or rear
r(trn) = radius of friction roller on trainer
m(b+r) = bike plus rider mass
m(fw) = mass of front wheel
m(fwe) = effective additional mass from front wheel rotation

You won't likely know the moment of inertia of the front wheel,
although it can be measured or possibly looked up. It won't be a big
contributor, and a rough estimate should be good enough. If all its
mass were at the radius, its additional effective mass from rotation
would equal its mass. If it were a uniform disc, its additional
effective mass would be 1/2 its mass. Using a number like 0.8 times
its mass should be close enough.

m(fwe) = J(fw)/[r(w)^2] = ~ 0.8 m(fw)

to match effective mass,

m(b+r) + m(fwe) = r(w)^2 * [J(trn) + J(fly)]/r(trn)^4

You probably won't know the moment of inertia of the trainer as-is.
If most of its moment of inertia comes from a flywheel, you can get
fairly close if you know the density of the flywheel material and
its geometry. For your added or replacement flywheel, you can keep
the geometry simple, such as a solid disc, from a known material like
steel.

As for the downhill, you won't get a match by just reducing the
flywheel, but that's about your only option. You could fudge it by
reducing the effective mass by your estimate of flatland acceleration/
downhill acceleration.

Dave Lehnen

On Jan 28, 10:14*pm, Dave Lehnen wrote:
Joseph S wrote:
Howdy All!

I'm deep in some winter training, and trying to do sprint workouts on
my new Kurt Kinetic Road Machine trainer. It's quite nice, but could
use a bit more inertia. There is an optional extra heavy flywheel
available, but they seem to be sold-out world wide, so I figure I
could get one turned out locally, and as a plus, I could have it's
mass tuned to suit me.

So I'm trying to figure out what flywheel weight and dimensions I
should spec. I'm most interested in sprint training, so I've been
trying to figure out my energy, but I'm not getting anywhere...

The Kurt does a good job of simulating wind resistance, so the trainer
speeds (and thus gearing) is similar to the real world, but the
trainer accelerates too fast.

Here is data for a typical (optimal) sprint I'd like to simulate:

Initial velocity: 35 km/h
Final velocity: 65 km/h
Duration: 10 sec
Distance: 150m
Total rider + bike mass: 115 kg

To complicate matters I do my sprints on a slight downhill that is
similar to the drop at a velodrome. About 5m.

I've guesstimated my acceleration (falsely assuming constant) to be
about .85 m/s/s ((18m/s - 9.7m/s) / 10s), and thus a force of about
98N. Given the 150m, that means 98*150= 14.7kJ. That makes sense. My
powertap tells me 13kJ but it doesn't know anything about the hill.

And that's where I get stuck. I can't figure out how to spec a
flywheel that can absorb that amount of energy at the appropriate
rate. Anybody feel like helping out?

Thanks,

Joseph

I'm going to ignore the hill, at least at first, and look at matching
effective mass. A passive trainer will never simulate a hill very well
in any case.

With the bike on the trainer, the total effective mass comes from the
rear wheel and drivetrain, the rider's legs, and from the moment of
inertia of the rotating parts of the trainer, including its flywheel.
When on the road, the total effective mass does not include anything
from the trainer, but adds the mass of the bike and rider, and the
effective mass from the moment of inertia of the front wheel. Since the
contribution from the rear wheel, drivetrain, and rider's legs is
identical in both cases, you don't need to worry about these. (The
contribution from effective moment of inertia of the legs would be a
complex calculation, but not a very big part of the total anyway.)

So to match effective mass, the total moment of inertia of the trainer
should make up for the rider mass, and bike mass (complete with both
wheels), and effective mass from the moment of inertia of the front
wheel.

The trainer spins faster than the bike wheel, by the ratio of the rear
tire diameter to the friction roller diameter on the trainer. The
effective mass from the trainer varies with the square of this ratio.

let:
J(fw) = moment of inertia of front wheel
J(trn) = moment of inertia of trainer before new flywheel
J(fly) = moment of inertia of added flywheel
r(w) = radius of wheel, front or rear
r(trn) = radius of friction roller on trainer
m(b+r) = bike plus rider mass
m(fw) = mass of front wheel
m(fwe) = effective additional mass from front wheel rotation

You won't likely know the moment of inertia of the front wheel,
although it can be measured or possibly looked up. It won't be a big
contributor, and a rough estimate should be good enough. If all its
mass were at the radius, its additional effective mass from rotation
would equal its mass. If it were a uniform disc, its additional
effective mass would be 1/2 its mass. Using a number like 0.8 times
its mass should be close enough.

m(fwe) = J(fw)/[r(w)^2] = ~ 0.8 m(fw)

to match effective mass,

m(b+r) + m(fwe) = r(w)^2 * [J(trn) + J(fly)]/r(trn)^4

You probably won't know the moment of inertia of the trainer as-is.
If most of its moment of inertia comes from a flywheel, you can get
fairly close if you know the density of the flywheel material and
its geometry. For your added or replacement flywheel, you can keep
the geometry simple, such as a solid disc, from a known material like
steel.

As for the downhill, you won't get a match by just reducing the
flywheel, but that's about your only option. You could fudge it by
reducing the effective mass by your estimate of flatland acceleration/
downhill acceleration.

Dave Lehnen

Outstanding! I knew there was a reason I wasn't able to figure it out
by myself ;-)

So I get a total desired flywheel inertia of .00966 and according to
the Kurt website, the stock FW has 2835g mass. I measured it and it is
a solid disc apparently of steel with radius 8cm, thickness 1.5cm
(roughly) which sounds about right for the mass (I think it's a bit
thicker than 1.5, but I just wanted to see if it was close before
digging out my calipers). So the inertia of the stock disk is about .
00907 (according to an online calc I found). That's about a 6.5%
difference.

So I wonder if perhaps I miscalculated my inertia, or if 6.5% is
actually more than it seems. It feels like it's more than 6.5%
difference. In real life were I to lose 6.5% of my mass, I still
wouldn't be able to accelerate from 30km/h to 60 km/h in one second
like I can on the Kurt.

The max practical diameter flywheel that could be fitted is about 22
cm. The roller is 54 mm. Once I get this figured out, I'm going to try
to simplify so I can say , "rider of X kg, needs flywheel of Y grams."

Joseph

On Jan 29, 1:08*pm, Joseph S wrote:
On Jan 28, 10:14*pm, Dave Lehnen wrote:





Joseph S wrote:
Howdy All!

I'm deep in some winter training, and trying to do sprint workouts on
my new Kurt Kinetic Road Machine trainer. It's quite nice, but could
use a bit more inertia. There is an optional extra heavy flywheel
available, but they seem to be sold-out world wide, so I figure I
could get one turned out locally, and as a plus, I could have it's
mass tuned to suit me.

So I'm trying to figure out what flywheel weight and dimensions I
should spec. I'm most interested in sprint training, so I've been
trying to figure out my energy, but I'm not getting anywhere...

The Kurt does a good job of simulating wind resistance, so the trainer
speeds (and thus gearing) is similar to the real world, but the
trainer accelerates too fast.

Here is data for a typical (optimal) sprint I'd like to simulate:

Initial velocity: 35 km/h
Final velocity: 65 km/h
Duration: 10 sec
Distance: 150m
Total rider + bike mass: 115 kg

To complicate matters I do my sprints on a slight downhill that is
similar to the drop at a velodrome. About 5m.

I've guesstimated my acceleration (falsely assuming constant) to be
about .85 m/s/s ((18m/s - 9.7m/s) / 10s), and thus a force of about
98N. Given the 150m, that means 98*150= 14.7kJ. That makes sense. My
powertap tells me 13kJ but it doesn't know anything about the hill.

And that's where I get stuck. I can't figure out how to spec a
flywheel that can absorb that amount of energy at the appropriate
rate. Anybody feel like helping out?

Thanks,

Joseph

I'm going to ignore the hill, at least at first, and look at matching
effective mass. A passive trainer will never simulate a hill very well
in any case.

With the bike on the trainer, the total effective mass comes from the
rear wheel and drivetrain, the rider's legs, and from the moment of
inertia of the rotating parts of the trainer, including its flywheel.
When on the road, the total effective mass does not include anything
from the trainer, but adds the mass of the bike and rider, and the
effective mass from the moment of inertia of the front wheel. Since the
contribution from the rear wheel, drivetrain, and rider's legs is
identical in both cases, you don't need to worry about these. (The
contribution from effective moment of inertia of the legs would be a
complex calculation, but not a very big part of the total anyway.)

So to match effective mass, the total moment of inertia of the trainer
should make up for the rider mass, and bike mass (complete with both
wheels), and effective mass from the moment of inertia of the front
wheel.

The trainer spins faster than the bike wheel, by the ratio of the rear
tire diameter to the friction roller diameter on the trainer. The
effective mass from the trainer varies with the square of this ratio.

let:
J(fw) = moment of inertia of front wheel
J(trn) = moment of inertia of trainer before new flywheel
J(fly) = moment of inertia of added flywheel
r(w) = radius of wheel, front or rear
r(trn) = radius of friction roller on trainer
m(b+r) = bike plus rider mass
m(fw) = mass of front wheel
m(fwe) = effective additional mass from front wheel rotation

You won't likely know the moment of inertia of the front wheel,
although it can be measured or possibly looked up. It won't be a big
contributor, and a rough estimate should be good enough. If all its
mass were at the radius, its additional effective mass from rotation
would equal its mass. If it were a uniform disc, its additional
effective mass would be 1/2 its mass. Using a number like 0.8 times
its mass should be close enough.

m(fwe) = J(fw)/[r(w)^2] = ~ 0.8 m(fw)

to match effective mass,

m(b+r) + m(fwe) = r(w)^2 * [J(trn) + J(fly)]/r(trn)^4

You probably won't know the moment of inertia of the trainer as-is.
If most of its moment of inertia comes from a flywheel, you can get
fairly close if you know the density of the flywheel material and
its geometry. For your added or replacement flywheel, you can keep
the geometry simple, such as a solid disc, from a known material like
steel.

As for the downhill, you won't get a match by just reducing the
flywheel, but that's about your only option. You could fudge it by
reducing the effective mass by your estimate of flatland acceleration/
downhill acceleration.

Dave Lehnen

Outstanding! I knew there was a reason I wasn't able to figure it out
by myself ;-)

So I get a total desired flywheel inertia of .00966 and according to
the Kurt website, the stock FW has 2835g mass. I measured it and it is
a solid disc apparently of steel with radius 8cm, thickness 1.5cm
(roughly) which sounds about right for the mass (I think it's a bit
thicker than 1.5, but I just wanted to see if it was close before
digging out my calipers). So the inertia of the stock disk is about .
00907 (according to an online calc I found). That's about a 6.5%
difference.

So I wonder if perhaps I miscalculated my inertia, or if 6.5% is
actually more than it seems. It feels like it's more than 6.5%
difference. In real life were I to lose 6.5% of my mass, I still
wouldn't be able to accelerate from 30km/h to 60 km/h in one second
like I can on the Kurt.

The max practical diameter flywheel that could be fitted is about 22
cm. The roller is 54 mm. Once I get this figured out, I'm going to try
to simplify so I can say , "rider of X kg, needs flywheel of Y grams."

Joseph

I just recalculated my desired total fw inertia to .00875 (being more
precise with my measurements of the roller and wheel diameter) and
realizing that small differences in the flywheel diameter make a big
difference, I measured and found the diameter to be 16.5cm or more
likely 6.5". That means the standard fw inertia is .00965 which is
MORE than I need. Now I'm confused...

Joseph

On Jan 29, 1:44*pm, Joseph S wrote:
On Jan 29, 1:08*pm, Joseph S wrote:





On Jan 28, 10:14*pm, Dave Lehnen wrote:

Joseph S wrote:
Howdy All!

I'm deep in some winter training, and trying to do sprint workouts on
my new Kurt Kinetic Road Machine trainer. It's quite nice, but could
use a bit more inertia. There is an optional extra heavy flywheel
available, but they seem to be sold-out world wide, so I figure I
could get one turned out locally, and as a plus, I could have it's
mass tuned to suit me.

So I'm trying to figure out what flywheel weight and dimensions I
should spec. I'm most interested in sprint training, so I've been
trying to figure out my energy, but I'm not getting anywhere...

The Kurt does a good job of simulating wind resistance, so the trainer
speeds (and thus gearing) is similar to the real world, but the
trainer accelerates too fast.

Here is data for a typical (optimal) sprint I'd like to simulate:

Initial velocity: 35 km/h
Final velocity: 65 km/h
Duration: 10 sec
Distance: 150m
Total rider + bike mass: 115 kg

To complicate matters I do my sprints on a slight downhill that is
similar to the drop at a velodrome. About 5m.

I've guesstimated my acceleration (falsely assuming constant) to be
about .85 m/s/s ((18m/s - 9.7m/s) / 10s), and thus a force of about
98N. Given the 150m, that means 98*150= 14.7kJ. That makes sense. My
powertap tells me 13kJ but it doesn't know anything about the hill.

And that's where I get stuck. I can't figure out how to spec a
flywheel that can absorb that amount of energy at the appropriate
rate. Anybody feel like helping out?

Thanks,

Joseph

I'm going to ignore the hill, at least at first, and look at matching
effective mass. A passive trainer will never simulate a hill very well
in any case.

With the bike on the trainer, the total effective mass comes from the
rear wheel and drivetrain, the rider's legs, and from the moment of
inertia of the rotating parts of the trainer, including its flywheel.
When on the road, the total effective mass does not include anything
from the trainer, but adds the mass of the bike and rider, and the
effective mass from the moment of inertia of the front wheel. Since the
contribution from the rear wheel, drivetrain, and rider's legs is
identical in both cases, you don't need to worry about these. (The
contribution from effective moment of inertia of the legs would be a
complex calculation, but not a very big part of the total anyway.)

So to match effective mass, the total moment of inertia of the trainer
should make up for the rider mass, and bike mass (complete with both
wheels), and effective mass from the moment of inertia of the front
wheel.

The trainer spins faster than the bike wheel, by the ratio of the rear
tire diameter to the friction roller diameter on the trainer. The
effective mass from the trainer varies with the square of this ratio.

let:
J(fw) = moment of inertia of front wheel
J(trn) = moment of inertia of trainer before new flywheel
J(fly) = moment of inertia of added flywheel
r(w) = radius of wheel, front or rear
r(trn) = radius of friction roller on trainer
m(b+r) = bike plus rider mass
m(fw) = mass of front wheel
m(fwe) = effective additional mass from front wheel rotation

You won't likely know the moment of inertia of the front wheel,
although it can be measured or possibly looked up. It won't be a big
contributor, and a rough estimate should be good enough. If all its
mass were at the radius, its additional effective mass from rotation
would equal its mass. If it were a uniform disc, its additional
effective mass would be 1/2 its mass. Using a number like 0.8 times
its mass should be close enough.

m(fwe) = J(fw)/[r(w)^2] = ~ 0.8 m(fw)

to match effective mass,

m(b+r) + m(fwe) = r(w)^2 * [J(trn) + J(fly)]/r(trn)^4

You probably won't know the moment of inertia of the trainer as-is.
If most of its moment of inertia comes from a flywheel, you can get
fairly close if you know the density of the flywheel material and
its geometry. For your added or replacement flywheel, you can keep
the geometry simple, such as a solid disc, from a known material like
steel.

As for the downhill, you won't get a match by just reducing the
flywheel, but that's about your only option. You could fudge it by
reducing the effective mass by your estimate of flatland acceleration/
downhill acceleration.

Dave Lehnen

Outstanding! I knew there was a reason I wasn't able to figure it out
by myself ;-)

So I get a total desired flywheel inertia of .00966 and according to
the Kurt website, the stock FW has 2835g mass. I measured it and it is
a solid disc apparently of steel with radius 8cm, thickness 1.5cm
(roughly) which sounds about right for the mass (I think it's a bit
thicker than 1.5, but I just wanted to see if it was close before
digging out my calipers). So the inertia of the stock disk is about .
00907 (according to an online calc I found). That's about a 6.5%
difference.

So I wonder if perhaps I miscalculated my inertia, or if 6.5% is
actually more than it seems. It feels like it's more than 6.5%
difference. In real life were I to lose 6.5% of my mass, I still
wouldn't be able to accelerate from 30km/h to 60 km/h in one second
like I can on the Kurt.

The max practical diameter flywheel that could be fitted is about 22
cm. The roller is 54 mm. Once I get this figured out, I'm going to try
to simplify so I can say , "rider of X kg, needs flywheel of Y grams."

Joseph

I just recalculated my desired total fw inertia to .00875 (being more
precise with my measurements of the roller and wheel diameter) and
realizing that small differences in the flywheel diameter make a big
difference, I measured and found the diameter to be 16.5cm or more
likely 6.5". That means the standard fw inertia is .00965 which is
MORE than I need. Now I'm confused...

Joseph

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

Joseph

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

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

Op 29-1-2011 18:03, Joseph S schreef:
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

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

Lou, bright sunshine at 1 degree Celcius today.

On Jan 29, 7:30*pm, Lou Holtman wrote:
Op 29-1-2011 18:03, Joseph S schreef:





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

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

Lou, bright sunshine at 1 degree Celcius today.

I did ride outside today! Bright sunshine, -6C. I use the Kurt for
sprint workouts, which I can't do on ice...

Joseph

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.

Chalo