some tire drop data

In article <[email hidden]>,

Quoted message said:

Some recent threads and an email today led me to measure how much a
fairly typical tire rose as I inflated it.

The results suggest that some people are fooling themselves.

First, some data from dial indicator measurements of a loaded tire's
height rise in thousandths of an inch from 40 to 130 psi, measured
from top of nominal 700x25c tire.

Tire measured 1.020" to 1.030" wide at top between 80 to 130 psi.

Total load on tire was ~85 pounds, resting on a scale, but the tire
stood on a concrete floor for the test.

inch mm inch
psi rise rise change
--- ----- ---- ------
40 0.000 0.00 n/a
50 0.021 0.53 0.021
60 0.038 0.97 0.017
70 0.052 1.32 0.014
80 0.063 1.60 0.011
90 0.076 1.93 0.013
100 0.088 2.24 0.012
110 0.100 2.54 0.012
120 0.110 2.79 0.010
130 0.120 3.05 0.010

My quick and dirty table shows only that the top of a roughly 1.020"
wide tire rose ~0.120 inches from 40 to 130 psi under an 88 pound
load.

For those interested in arbitrary figures, 15% of a tire width of
1.020" is 0.153", or 3.9 mm.

Presumably, someone could measure the diameter of a tire hanging in
mid-air and inflated to some arbitrary pressure, load it with some
weights, measure the change, and call it tire drop.

But this would require such an elaborate test rig that it would be
extremely difficult for any normal poster to even _try_ to make a
practical measurement of the difference between 90 psi and 100 psi.

After all, the difference between 90 and 100 psi with an elaborate
test rig using a dial indicator was only 0.012", about a hundredth of
an inch.

Incidentally, it is highly unlikely that pressure and a 700x25 tire's
rise have the straight linear relationship that some people suggest,
at least not over a 40 to 130 psi range with an 88 pound load.

I expect that beautifully smooth graphs of pressure and rise are just
as theoretical, oversimplified, and mistaken as similar predictions
for contact patch area, which does not follow the often-mentioned
tire-pressure/load = area prediction.

(Actual measurements of contact patches show that they fail to shrink
as much as predicted with higher pressures and fail to expand as much
as predicted with lower pressures. They stubbornly stick to a
preferred size.)

Interesting data from Fogel Labs again; thanks, Carl. I don't have the
original article about the 15% drop figure right here to read again.
IIRC the 15% drop figure came from research by Jim Papodopolous, but how
it was arrived at I don't know. It's a number that looks suspiciously
arbitrary (how many things in the world work out to such a nice even
number?).

Tim McNamara said:

In article <[email hidden]>,

Quoted message said:

Some recent threads and an email today led me to measure how much a
fairly typical tire rose as I inflated it.

The results suggest that some people are fooling themselves.

First, some data from dial indicator measurements of a loaded tire's
height rise in thousandths of an inch from 40 to 130 psi, measured
from top of nominal 700x25c tire.

Tire measured 1.020" to 1.030" wide at top between 80 to 130 psi.

Total load on tire was ~85 pounds, resting on a scale, but the tire
stood on a concrete floor for the test.

inch mm inch
psi rise rise change
--- ----- ---- ------
40 0.000 0.00 n/a
50 0.021 0.53 0.021
60 0.038 0.97 0.017
70 0.052 1.32 0.014
80 0.063 1.60 0.011
90 0.076 1.93 0.013
100 0.088 2.24 0.012
110 0.100 2.54 0.012
120 0.110 2.79 0.010
130 0.120 3.05 0.010

My quick and dirty table shows only that the top of a roughly 1.020"
wide tire rose ~0.120 inches from 40 to 130 psi under an 88 pound
load.

For those interested in arbitrary figures, 15% of a tire width of
1.020" is 0.153", or 3.9 mm.

Presumably, someone could measure the diameter of a tire hanging in
mid-air and inflated to some arbitrary pressure, load it with some
weights, measure the change, and call it tire drop.

But this would require such an elaborate test rig that it would be
extremely difficult for any normal poster to even _try_ to make a
practical measurement of the difference between 90 psi and 100 psi.

After all, the difference between 90 and 100 psi with an elaborate
test rig using a dial indicator was only 0.012", about a hundredth of
an inch.

Incidentally, it is highly unlikely that pressure and a 700x25 tire's
rise have the straight linear relationship that some people suggest,
at least not over a 40 to 130 psi range with an 88 pound load.

I expect that beautifully smooth graphs of pressure and rise are just
as theoretical, oversimplified, and mistaken as similar predictions
for contact patch area, which does not follow the often-mentioned
tire-pressure/load = area prediction.

(Actual measurements of contact patches show that they fail to shrink
as much as predicted with higher pressures and fail to expand as much
as predicted with lower pressures. They stubbornly stick to a
preferred size.)

Interesting data from Fogel Labs again; thanks, Carl. I don't have the
original article about the 15% drop figure right here to read again.
IIRC the 15% drop figure came from research by Jim Papodopolous, but how
it was arrived at I don't know. It's a number that looks suspiciously
arbitrary (how many things in the world work out to such a nice even
number?).

Dear Tim,

I don't know where the popular 15% figure came from or from what point
it was measured--just the weight of the front end of a bike resting on
the tire is about 10% of the typical load.

Maybe there's a clever measuring technique that was somehow confirmed
by rolling resistance tests, but it's going to have to be awfully
clever to handle differences of about a hundredth of an inch between
90 and 100 psi.

Here's a link that came in my email to a chart that graphs some kind
of 15% drop correlated with pressure for various tire sizes:

http://bp1.blogger.com/_d-Yj0VDKhWQ/RnwLUoA9sVI/AAAAAAAAAJ4/225Ym5qYKTo/s1600-h/BQ_berto_inflationgraph.jpg

The graph seems to show some simple linear calculations, not actual
data points from measurements.

I suspect that actual measurements would show the straight line theory
is simply wrong.

If a tire were a piston in a car engine, then adding a 100 pound
weight (including piston) would drive a piston with an area of 1
square inch down until the pressure rose to 100 psi.

But a bicycle tire is significantly different from a metal piston in a
metal chamber.

When we load a tire, the bottom of the "piston chamber" is the
reasonably unyielding metal rim.

But on either side of the tire, the C-shaped tire walls deform outward
against the air pressure trying to keep them in their current shape
about an inch wide on a 25 mm tire, the minor axis of the toroid (your
thumb and finger going around the tire).

Meanwhile, the tire also deforms, much less against the gigantic "O"
of the tire viewed the long way around, the major axis of the toroid.

In effect, the tire is more like a long, narrow trampoline than the
top of a metal piston, and it distorts in a much more complicated
fashion.

This is why the simple area = load / pressure load equation fails to
predict what happens at high and low pressures for 700c x 25 tires.

The same explanation probably applies to how high a tire rises or
falls according to pressure. The sides of the tire act as scissors
jacks, applying increasing pressure at the edge of the contact patch
as pressure drops and the sides deform more and more.

As I've mentioned before, what made me suspicious of simple
straight-line predictions was knowing that a 4-inch wide trials tire
can carry a 400-pound load in a wheelstand down a paved road with only
4 psi and no way to spread out more than 4 inches. The oversimplified
equation predicts:

400 lbs / 4 (lbs/inch^2) = 100 ^ 2 = 4 x 25 inches

But a 4 x 18 trials tire is only about 26 inches in diameter, so a 25
inch long contact patch is impossible. The sidewalls had to be
exerting considerable force as springs, which seems reasonable, since
it takes considerable force to pull (or push) the wall of an inflated
tire out of shape.

Cheers,

Carl Fogel

Tim McNamara said:

In article <[email hidden]>,

Quoted message said:

Some recent threads and an email today led me to measure how much a
fairly typical tire rose as I inflated it.

Quoted message said:

The results suggest that some people are fooling themselves.

Quoted message said:

First, some data from dial indicator measurements of a loaded tire's
height rise in thousandths of an inch from 40 to 130 psi, measured
from top of nominal 700x25c tire.

Quoted message said:

Tire measured 1.020" to 1.030" wide at top between 80 to 130 psi.

Quoted message said:

Total load on tire was ~85 pounds, resting on a scale, but the tire
stood on a concrete floor for the test.

Quoted message said:

inch mm inch
psi rise rise change
--- ----- ---- ------
40 0.000 0.00 n/a
50 0.021 0.53 0.021
60 0.038 0.97 0.017
70 0.052 1.32 0.014
80 0.063 1.60 0.011
90 0.076 1.93 0.013
100 0.088 2.24 0.012
110 0.100 2.54 0.012
120 0.110 2.79 0.010
130 0.120 3.05 0.010

Quoted message said:

My quick and dirty table shows only that the top of a roughly 1.020"
wide tire rose ~0.120 inches from 40 to 130 psi under an 88 pound
load.

Quoted message said:

For those interested in arbitrary figures, 15% of a tire width of
1.020" is 0.153", or 3.9 mm.

Quoted message said:

Presumably, someone could measure the diameter of a tire hanging in
mid-air and inflated to some arbitrary pressure, load it with some
weights, measure the change, and call it tire drop.

Quoted message said:

But this would require such an elaborate test rig that it would be
extremely difficult for any normal poster to even _try_ to make a
practical measurement of the difference between 90 psi and 100 psi.

Quoted message said:

After all, the difference between 90 and 100 psi with an elaborate
test rig using a dial indicator was only 0.012", about a hundredth of
an inch.

Quoted message said:

Incidentally, it is highly unlikely that pressure and a 700x25 tire's
rise have the straight linear relationship that some people suggest,
at least not over a 40 to 130 psi range with an 88 pound load.

Quoted message said:

I expect that beautifully smooth graphs of pressure and rise are just
as theoretical, oversimplified, and mistaken as similar predictions
for contact patch area, which does not follow the often-mentioned
tire-pressure/load = area prediction.

Quoted message said:

(Actual measurements of contact patches show that they fail to shrink
as much as predicted with higher pressures and fail to expand as much
as predicted with lower pressures. They stubbornly stick to a
preferred size.)

Interesting data from Fogel Labs again; thanks, Carl. I don't have the
original article about the 15% drop figure right here to read again.
IIRC the 15% drop figure came from research by Jim Papodopolous, but how
it was arrived at I don't know. It's a number that looks suspiciously
arbitrary (how many things in the world work out to such a nice even
number?).- Hide quoted text -

- Show quoted text -

Thanks for the information, Carl. I think the original article came
from Bicycling magazine. Currently you can get a version of it on the
roadbikerider.com website in the member's section.

Smokey

"Tim McNamara" <[email hidden]> wrote in message
news:[email hidden]...
| In article <[email hidden]>,
| [email hidden] wrote:
|
| > Some recent threads and an email today led me to measure how much a
| > fairly typical tire rose as I inflated it.
| >
| > The results suggest that some people are fooling themselves.
| >
.....snip
|
| Interesting data from Fogel Labs again; thanks, Carl. I don't have the
| original article about the 15% drop figure right here to read again.
| IIRC the 15% drop figure came from research by Jim Papodopolous, but how
| it was arrived at I don't know. It's a number that looks suspiciously
| arbitrary (how many things in the world work out to such a nice even
| number?).

I don't know how Berto came up with the 15% figure, but I do have a
clipping from Bicycling magazine that says "Michelin and National recommend
selecting tire size and inflation pressure so the tire height drops by 20%
when you get on the bike". How they arrived at the 20% figure was never
mentioned.

In article <[email hidden]>,

Quoted message said:

Tim McNamara said:

In article <[email hidden]>,

Quoted message said:

Some recent threads and an email today led me to measure how much
a fairly typical tire rose as I inflated it.

The results suggest that some people are fooling themselves.

First, some data from dial indicator measurements of a loaded
tire's height rise in thousandths of an inch from 40 to 130 psi,
measured from top of nominal 700x25c tire.

Tire measured 1.020" to 1.030" wide at top between 80 to 130 psi.

Total load on tire was ~85 pounds, resting on a scale, but the
tire stood on a concrete floor for the test.

inch mm inch
psi rise rise change
--- ----- ---- ------
40 0.000 0.00 n/a
50 0.021 0.53 0.021
60 0.038 0.97 0.017
70 0.052 1.32 0.014
80 0.063 1.60 0.011
90 0.076 1.93 0.013
100 0.088 2.24 0.012
110 0.100 2.54 0.012
120 0.110 2.79 0.010
130 0.120 3.05 0.010

My quick and dirty table shows only that the top of a roughly
1.020" wide tire rose ~0.120 inches from 40 to 130 psi under an 88
pound load.

For those interested in arbitrary figures, 15% of a tire width of
1.020" is 0.153", or 3.9 mm.

Presumably, someone could measure the diameter of a tire hanging
in mid-air and inflated to some arbitrary pressure, load it with
some weights, measure the change, and call it tire drop.

But this would require such an elaborate test rig that it would be
extremely difficult for any normal poster to even _try_ to make a
practical measurement of the difference between 90 psi and 100
psi.

After all, the difference between 90 and 100 psi with an elaborate
test rig using a dial indicator was only 0.012", about a hundredth
of an inch.

Incidentally, it is highly unlikely that pressure and a 700x25
tire's rise have the straight linear relationship that some people
suggest, at least not over a 40 to 130 psi range with an 88 pound
load.

I expect that beautifully smooth graphs of pressure and rise are
just as theoretical, oversimplified, and mistaken as similar
predictions for contact patch area, which does not follow the
often-mentioned tire-pressure/load = area prediction.

(Actual measurements of contact patches show that they fail to
shrink as much as predicted with higher pressures and fail to
expand as much as predicted with lower pressures. They stubbornly
stick to a preferred size.)

Interesting data from Fogel Labs again; thanks, Carl. I don't have
the original article about the 15% drop figure right here to read
again. IIRC the 15% drop figure came from research by Jim
Papodopolous, but how it was arrived at I don't know. It's a number
that looks suspiciously arbitrary (how many things in the world work
out to such a nice even number?).

Dear Tim,

I don't know where the popular 15% figure came from or from what
point it was measured--just the weight of the front end of a bike
resting on the tire is about 10% of the typical load.

Having reviewed the article in Bicycle Quarterly ("Optimizing your tire
pressure for your weight," BQ Vol 5, Issue 4), the 15% figure was
referenced from an article by Frank Berto (2004. Under Pressure.
Australian Cyclist March/April 2004, p. 48). I was not able to locate
the article on the Australian Cyclist Web site.

Heine measure loads on three bicycles with a rider seated on the bike
(randonneur bike, racing bike and "city" bike) and found that the weight
distribution (% front/rear) was 45/55, 30/60 and 35/65 respectively.

Heine's article references that a 15% "tire drop" is recommended by
"several tire manufacturers." That statement is cited to Berto's
original article and no more specifics are given. The chart with the
nice linear relationships is also attributed to Berto.

Quoted message said:

Maybe there's a clever measuring technique that was somehow confirmed
by rolling resistance tests, but it's going to have to be awfully
clever to handle differences of about a hundredth of an inch between
90 and 100 psi.

My guess would be to load the wheel through a frame or fork and measure
the drop from the top of the tire to the ground, with an arrangement
similar to how a doctor measures your height. It would indeed need to
be carefully calibrated, perhaps some dial arrangement could be used.

Quoted message said:

Here's a link that came in my email to a chart that graphs some kind
of 15% drop correlated with pressure for various tire sizes:

http://bp1.blogger.com/_d-Yj0VDKhWQ/RnwLUoA9sVI/AAAAAAAAAJ4/225Ym5qYKT
o/s1600- h/BQ_berto_inflationgraph.jpg

The graph seems to show some simple linear calculations, not actual
data points from measurements.

I suspect that actual measurements would show the straight line
theory is simply wrong.

That is the graph included on p 29 of BQ 5,4. It is attributed to
Berto's article in Australian Cyclist. Without being able to read the
Berto article, there isn't any way to understand how Berto came to
create that graph. I agree that such a linear relationship between
inflation and tire drop is probably unlikely, although it's conceivable
that the relationship might be close to linear within the normal range
in inflation pressures. If you look at the Avocet rolling resistance
data, the curve of the relationship between inflation pressure and
rolling resistance flattened quite a bit at higher pressures.

Maybe you can Google up the source article by Berto, or perhaps there is
some Australian or New Zealand or Tasmanian participant in the newsgroup
who can provide that information.

"Tim McNamara" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

In article <[email hidden]>,

Quoted message said:

Tim McNamara said:

In article <[email hidden]>,
[email hidden] wrote:

> Some recent threads and an email today led me to measure how much
> a fairly typical tire rose as I inflated it.
>
> The results suggest that some people are fooling themselves.
>
> First, some data from dial indicator measurements of a loaded
> tire's height rise in thousandths of an inch from 40 to 130 psi,
> measured from top of nominal 700x25c tire.
>
> Tire measured 1.020" to 1.030" wide at top between 80 to 130 psi.
>
> Total load on tire was ~85 pounds, resting on a scale, but the
> tire stood on a concrete floor for the test.
>
> inch mm inch
> psi rise rise change
> --- ----- ---- ------
> 40 0.000 0.00 n/a
> 50 0.021 0.53 0.021
> 60 0.038 0.97 0.017
> 70 0.052 1.32 0.014
> 80 0.063 1.60 0.011
> 90 0.076 1.93 0.013
> 100 0.088 2.24 0.012
> 110 0.100 2.54 0.012
> 120 0.110 2.79 0.010
> 130 0.120 3.05 0.010
>
> My quick and dirty table shows only that the top of a roughly
> 1.020" wide tire rose ~0.120 inches from 40 to 130 psi under an 88
> pound load.
>
> For those interested in arbitrary figures, 15% of a tire width of
> 1.020" is 0.153", or 3.9 mm.
>
> Presumably, someone could measure the diameter of a tire hanging
> in mid-air and inflated to some arbitrary pressure, load it with
> some weights, measure the change, and call it tire drop.
>
> But this would require such an elaborate test rig that it would be
> extremely difficult for any normal poster to even _try_ to make a
> practical measurement of the difference between 90 psi and 100
> psi.
>
> After all, the difference between 90 and 100 psi with an elaborate
> test rig using a dial indicator was only 0.012", about a hundredth
> of an inch.
>
> Incidentally, it is highly unlikely that pressure and a 700x25
> tire's rise have the straight linear relationship that some people
> suggest, at least not over a 40 to 130 psi range with an 88 pound
> load.
>
> I expect that beautifully smooth graphs of pressure and rise are
> just as theoretical, oversimplified, and mistaken as similar
> predictions for contact patch area, which does not follow the
> often-mentioned tire-pressure/load = area prediction.
>
> (Actual measurements of contact patches show that they fail to
> shrink as much as predicted with higher pressures and fail to
> expand as much as predicted with lower pressures. They stubbornly
> stick to a preferred size.)

Interesting data from Fogel Labs again; thanks, Carl. I don't have
the original article about the 15% drop figure right here to read
again. IIRC the 15% drop figure came from research by Jim
Papodopolous, but how it was arrived at I don't know. It's a number
that looks suspiciously arbitrary (how many things in the world work
out to such a nice even number?).

Dear Tim,

I don't know where the popular 15% figure came from or from what
point it was measured--just the weight of the front end of a bike
resting on the tire is about 10% of the typical load.

Having reviewed the article in Bicycle Quarterly ("Optimizing your tire
pressure for your weight," BQ Vol 5, Issue 4), the 15% figure was
referenced from an article by Frank Berto (2004. Under Pressure.
Australian Cyclist March/April 2004, p. 48). I was not able to locate
the article on the Australian Cyclist Web site.

Heine measure loads on three bicycles with a rider seated on the bike
(randonneur bike, racing bike and "city" bike) and found that the weight
distribution (% front/rear) was 45/55, 30/60 and 35/65 respectively.

Heine's article references that a 15% "tire drop" is recommended by
"several tire manufacturers." That statement is cited to Berto's
original article and no more specifics are given. The chart with the
nice linear relationships is also attributed to Berto.

Quoted message said:

Maybe there's a clever measuring technique that was somehow confirmed
by rolling resistance tests, but it's going to have to be awfully
clever to handle differences of about a hundredth of an inch between
90 and 100 psi.

My guess would be to load the wheel through a frame or fork and measure
the drop from the top of the tire to the ground, with an arrangement
similar to how a doctor measures your height. It would indeed need to
be carefully calibrated, perhaps some dial arrangement could be used.

Quoted message said:

Here's a link that came in my email to a chart that graphs some kind
of 15% drop correlated with pressure for various tire sizes:

http://bp1.blogger.com/_d-Yj0VDKhWQ/RnwLUoA9sVI/AAAAAAAAAJ4/225Ym5qYKT
o/s1600- h/BQ_berto_inflationgraph.jpg

The graph seems to show some simple linear calculations, not actual
data points from measurements.

I suspect that actual measurements would show the straight line
theory is simply wrong.

That is the graph included on p 29 of BQ 5,4. It is attributed to
Berto's article in Australian Cyclist. Without being able to read the
Berto article, there isn't any way to understand how Berto came to
create that graph. I agree that such a linear relationship between
inflation and tire drop is probably unlikely, although it's conceivable
that the relationship might be close to linear within the normal range
in inflation pressures. If you look at the Avocet rolling resistance
data, the curve of the relationship between inflation pressure and
rolling resistance flattened quite a bit at higher pressures.

Maybe you can Google up the source article by Berto, or perhaps there is
some Australian or New Zealand or Tasmanian participant in the newsgroup
who can provide that information.

Years ago I remember seeing a study that showed only a 5%-7% reduction in
rolling resistance between 85 PSI and 110 PSI in a variety of tires.

Since most of us rode sewups in goathead country we opted to run at 85 PSI
to 90 PSI and got a lot less flats ( refer to Carl Fogel's test of tire
puncture vs. pressure ).

Chas.

carlfogel said:

Tim McNamara said:

In article <[email hidden]>,

Quoted message said:

Some recent threads and an email today led me to measure how much a
fairly typical tire rose as I inflated it.

The results suggest that some people are fooling themselves.

First, some data from dial indicator measurements of a loaded tire's
height rise in thousandths of an inch from 40 to 130 psi, measured
from top of nominal 700x25c tire.

Tire measured 1.020" to 1.030" wide at top between 80 to 130 psi.

Total load on tire was ~85 pounds, resting on a scale, but the tire
stood on a concrete floor for the test.

inch mm inch
psi rise rise change
--- ----- ---- ------
40 0.000 0.00 n/a
50 0.021 0.53 0.021
60 0.038 0.97 0.017
70 0.052 1.32 0.014
80 0.063 1.60 0.011
90 0.076 1.93 0.013
100 0.088 2.24 0.012
110 0.100 2.54 0.012
120 0.110 2.79 0.010
130 0.120 3.05 0.010

My quick and dirty table shows only that the top of a roughly 1.020"
wide tire rose ~0.120 inches from 40 to 130 psi under an 88 pound
load.

For those interested in arbitrary figures, 15% of a tire width of
1.020" is 0.153", or 3.9 mm.

Presumably, someone could measure the diameter of a tire hanging in
mid-air and inflated to some arbitrary pressure, load it with some
weights, measure the change, and call it tire drop.

But this would require such an elaborate test rig that it would be
extremely difficult for any normal poster to even _try_ to make a
practical measurement of the difference between 90 psi and 100 psi.

After all, the difference between 90 and 100 psi with an elaborate
test rig using a dial indicator was only 0.012", about a hundredth of
an inch.

Incidentally, it is highly unlikely that pressure and a 700x25 tire's
rise have the straight linear relationship that some people suggest,
at least not over a 40 to 130 psi range with an 88 pound load.

I expect that beautifully smooth graphs of pressure and rise are just
as theoretical, oversimplified, and mistaken as similar predictions
for contact patch area, which does not follow the often-mentioned
tire-pressure/load = area prediction.

(Actual measurements of contact patches show that they fail to shrink
as much as predicted with higher pressures and fail to expand as much
as predicted with lower pressures. They stubbornly stick to a
preferred size.)

Interesting data from Fogel Labs again; thanks, Carl. I don't have the
original article about the 15% drop figure right here to read again.
IIRC the 15% drop figure came from research by Jim Papodopolous, but how
it was arrived at I don't know. It's a number that looks suspiciously
arbitrary (how many things in the world work out to such a nice even
number?).

Dear Tim,

I don't know where the popular 15% figure came from or from what point
it was measured--just the weight of the front end of a bike resting on
the tire is about 10% of the typical load.

Maybe there's a clever measuring technique that was somehow confirmed
by rolling resistance tests, but it's going to have to be awfully
clever to handle differences of about a hundredth of an inch between
90 and 100 psi.

Here's a link that came in my email to a chart that graphs some kind
of 15% drop correlated with pressure for various tire sizes:

http://bp1.blogger.com/_d-Yj0VDKhWQ/RnwLUoA9sVI/AAAAAAAAAJ4/225Ym5qYKTo/s1600-h/BQ_berto_inflationgraph.jpg

The graph seems to show some simple linear calculations, not actual
data points from measurements.

I suspect that actual measurements would show the straight line theory
is simply wrong.

If a tire were a piston in a car engine, then adding a 100 pound
weight (including piston) would drive a piston with an area of 1
square inch down until the pressure rose to 100 psi.

But a bicycle tire is significantly different from a metal piston in a
metal chamber.

When we load a tire, the bottom of the "piston chamber" is the
reasonably unyielding metal rim.

But on either side of the tire, the C-shaped tire walls deform outward
against the air pressure trying to keep them in their current shape
about an inch wide on a 25 mm tire, the minor axis of the toroid (your
thumb and finger going around the tire).

Meanwhile, the tire also deforms, much less against the gigantic "O"
of the tire viewed the long way around, the major axis of the toroid.

In effect, the tire is more like a long, narrow trampoline than the
top of a metal piston, and it distorts in a much more complicated
fashion.

This is why the simple area = load / pressure load equation fails to
predict what happens at high and low pressures for 700c x 25 tires.

The same explanation probably applies to how high a tire rises or
falls according to pressure. The sides of the tire act as scissors
jacks, applying increasing pressure at the edge of the contact patch
as pressure drops and the sides deform more and more.

As I've mentioned before, what made me suspicious of simple
straight-line predictions was knowing that a 4-inch wide trials tire
can carry a 400-pound load in a wheelstand down a paved road with only
4 psi and no way to spread out more than 4 inches. The oversimplified
equation predicts:

400 lbs / 4 (lbs/inch^2) = 100 ^ 2 = 4 x 25 inches

But a 4 x 18 trials tire is only about 26 inches in diameter, so a 25
inch long contact patch is impossible. The sidewalls had to be
exerting considerable force as springs, which seems reasonable, since
it takes considerable force to pull (or push) the wall of an inflated
tire out of shape.

Cheers,

Carl Fogel

Jan Heine, publisher of the magazine where the graph appeared, posted a
response on the ibob mailing list:

"The graph with the linear relationships quoted in the link below
[i.e., Carl's link, above] came from Bicycle Quarterly. It was based on
actual measurements by Frank Berto, and his measurements did line up
roughly on a straight line (I have seen the original measurements). He
built a test rig, which he described many years ago in Bicycling Magazine.
The graphs are extrapolated toward the ends...

The discrepancy between Carl Fogel's measurements and Frank Berto's
might be easy to explain. They measured different things. Carl
measured the rise of the tire with increasing pressure. I am not
surprised that this is not linear - the tire cannot expand
indefinitely.

Frank Berto measured how much pressure he needed for different
weights to achieve the same tire deflection of 15%. The 15% came from
tire manufacturers' recommendations. In Bicycle Quarterly's real road
tests, we found that performance decreased significantly once the
"tire drop" was more than about 13-15%. With less tire drop, higher
pressures did not yield significant performance benefits.

Please feel free to forward this clarification to rec.bike.tech.

Jan Heine
Editor
Bicycle Quarterly
140 Lakeside Ave #C
Seattle WA 98122
www.bikequarterly.com"

http://search.bikelist.org/getmsg.asp?Filename=internet-bob.10709.1869.eml

On Sat, 29 Sep 2007 12:23:39 -0500, Tim McNamara
<[email hidden]> wrote:

[snip]

Quoted message said:

Quoted message said:

Maybe there's a clever measuring technique that was somehow confirmed
by rolling resistance tests, but it's going to have to be awfully
clever to handle differences of about a hundredth of an inch between
90 and 100 psi.

My guess would be to load the wheel through a frame or fork and measure
the drop from the top of the tire to the ground, with an arrangement
similar to how a doctor measures your height. It would indeed need to
be carefully calibrated, perhaps some dial arrangement could be used.

[snip]

Dear Tim,

In this case, the more careful the measurement, the sillier the
results will be. The resolution is already too fine for the subject.

There's no medical point, for example, to weighing normal adults to
within even a pound, since our weight easily varies that much
according to our last meal and visit to the bathroom. (The doctor does
not bother to have you remove your wallet, much less strip naked, when
he weighs you on his balance scales, accurate to 4 ounces.)

The "15%" tire drop figure implies resolution to at least 5%
increments, meaning that we think that 15% is what we want, not 10% or
20%.

But when compared to 10% or 20% drops, a "15% drop" means only that
somewhere between 70 and 110 psi is a good idea for tire inflation.

We hardly need elaborate charts and dial indicators to tell us that.

Here's how it works.

If a 1.000" wide tire has an absolute 15% drop of 0.150" at 90 psi
under an 85-lb load (don't ask me how to actually measure that with
repeatable accuracy), then we can apply my relative rise measurements
for a ~1" tire under the same conditions:

10 psi
theoretical measured
absolute relative
psi drop drops
--- -------- --------
x 20% 0.200" 20%
130 0.194" +0.010"
120 0.184" +0.010"
110 0.174" +0.012" \
100 0.162" +0.012" \
90 15% 0.150" 0 > 15% range
80 0.137" -0.013" /
70 0.126" -0.011" /
60 0.112" -0.014"
x 10% 0.100" 10%
50 0.091" -0.021"

Here's my test data again:

inch inch
psi rise change
--- ----- ------
40 0.000 n/a
50 0.021 0.021
60 0.038 0.017
70 0.052 0.014
80 0.063 0.011
90 0.076 0.013
100 0.088 0.012
110 0.100 0.012
120 0.110 0.010
130 0.120 0.010

In any case, inflated tires are not round to within a hundredth of an
inch. They're just slapped on aluminum rims and inflated, with the
beads creeping out roughly regularly, and then the sides bend as the
tire is loaded.

What's bending in clinchers are bias-laid plies of threads, with 66 to
170 threads per inch:

66 tpi 0.015"
127 tpi 0.008"
170 tpi 0.006"

A single thread is roughly as thick as a measured 10 psi difference in
tire drop.

Cheers,

Carl Fogel

On Sat, 29 Sep 2007 14:23:23 -0500, Gary Young <[email hidden]>

Quoted message said:

carlfogel said:

Tim McNamara said:

In article <[email hidden]>,
[email hidden] wrote:

> Some recent threads and an email today led me to measure how much a
> fairly typical tire rose as I inflated it.
>
> The results suggest that some people are fooling themselves.
>
> First, some data from dial indicator measurements of a loaded tire's
> height rise in thousandths of an inch from 40 to 130 psi, measured
> from top of nominal 700x25c tire.
>
> Tire measured 1.020" to 1.030" wide at top between 80 to 130 psi.
>
> Total load on tire was ~85 pounds, resting on a scale, but the tire
> stood on a concrete floor for the test.
>
> inch mm inch
> psi rise rise change
> --- ----- ---- ------
> 40 0.000 0.00 n/a
> 50 0.021 0.53 0.021
> 60 0.038 0.97 0.017
> 70 0.052 1.32 0.014
> 80 0.063 1.60 0.011
> 90 0.076 1.93 0.013
> 100 0.088 2.24 0.012
> 110 0.100 2.54 0.012
> 120 0.110 2.79 0.010
> 130 0.120 3.05 0.010
>
> My quick and dirty table shows only that the top of a roughly 1.020"
> wide tire rose ~0.120 inches from 40 to 130 psi under an 88 pound
> load.
>
> For those interested in arbitrary figures, 15% of a tire width of
> 1.020" is 0.153", or 3.9 mm.
>
> Presumably, someone could measure the diameter of a tire hanging in
> mid-air and inflated to some arbitrary pressure, load it with some
> weights, measure the change, and call it tire drop.
>
> But this would require such an elaborate test rig that it would be
> extremely difficult for any normal poster to even _try_ to make a
> practical measurement of the difference between 90 psi and 100 psi.
>
> After all, the difference between 90 and 100 psi with an elaborate
> test rig using a dial indicator was only 0.012", about a hundredth of
> an inch.
>
> Incidentally, it is highly unlikely that pressure and a 700x25 tire's
> rise have the straight linear relationship that some people suggest,
> at least not over a 40 to 130 psi range with an 88 pound load.
>
> I expect that beautifully smooth graphs of pressure and rise are just
> as theoretical, oversimplified, and mistaken as similar predictions
> for contact patch area, which does not follow the often-mentioned
> tire-pressure/load = area prediction.
>
> (Actual measurements of contact patches show that they fail to shrink
> as much as predicted with higher pressures and fail to expand as much
> as predicted with lower pressures. They stubbornly stick to a
> preferred size.)

Interesting data from Fogel Labs again; thanks, Carl. I don't have the
original article about the 15% drop figure right here to read again.
IIRC the 15% drop figure came from research by Jim Papodopolous, but how
it was arrived at I don't know. It's a number that looks suspiciously
arbitrary (how many things in the world work out to such a nice even
number?).

Dear Tim,

I don't know where the popular 15% figure came from or from what point
it was measured--just the weight of the front end of a bike resting on
the tire is about 10% of the typical load.

Maybe there's a clever measuring technique that was somehow confirmed
by rolling resistance tests, but it's going to have to be awfully
clever to handle differences of about a hundredth of an inch between
90 and 100 psi.

Here's a link that came in my email to a chart that graphs some kind
of 15% drop correlated with pressure for various tire sizes:

http://bp1.blogger.com/_d-Yj0VDKhWQ/RnwLUoA9sVI/AAAAAAAAAJ4/225Ym5qYKTo/s1600-h/BQ_berto_inflationgraph.jpg

The graph seems to show some simple linear calculations, not actual
data points from measurements.

I suspect that actual measurements would show the straight line theory
is simply wrong.

If a tire were a piston in a car engine, then adding a 100 pound
weight (including piston) would drive a piston with an area of 1
square inch down until the pressure rose to 100 psi.

But a bicycle tire is significantly different from a metal piston in a
metal chamber.

When we load a tire, the bottom of the "piston chamber" is the
reasonably unyielding metal rim.

But on either side of the tire, the C-shaped tire walls deform outward
against the air pressure trying to keep them in their current shape
about an inch wide on a 25 mm tire, the minor axis of the toroid (your
thumb and finger going around the tire).

Meanwhile, the tire also deforms, much less against the gigantic "O"
of the tire viewed the long way around, the major axis of the toroid.

In effect, the tire is more like a long, narrow trampoline than the
top of a metal piston, and it distorts in a much more complicated
fashion.

This is why the simple area = load / pressure load equation fails to
predict what happens at high and low pressures for 700c x 25 tires.

The same explanation probably applies to how high a tire rises or
falls according to pressure. The sides of the tire act as scissors
jacks, applying increasing pressure at the edge of the contact patch
as pressure drops and the sides deform more and more.

As I've mentioned before, what made me suspicious of simple
straight-line predictions was knowing that a 4-inch wide trials tire
can carry a 400-pound load in a wheelstand down a paved road with only
4 psi and no way to spread out more than 4 inches. The oversimplified
equation predicts:

400 lbs / 4 (lbs/inch^2) = 100 ^ 2 = 4 x 25 inches

But a 4 x 18 trials tire is only about 26 inches in diameter, so a 25
inch long contact patch is impossible. The sidewalls had to be
exerting considerable force as springs, which seems reasonable, since
it takes considerable force to pull (or push) the wall of an inflated
tire out of shape.

Cheers,

Carl Fogel

Jan Heine, publisher of the magazine where the graph appeared, posted a
response on the ibob mailing list:

"The graph with the linear relationships quoted in the link below
[i.e., Carl's link, above] came from Bicycle Quarterly. It was based on
actual measurements by Frank Berto, and his measurements did line up
roughly on a straight line (I have seen the original measurements). He
built a test rig, which he described many years ago in Bicycling Magazine.
The graphs are extrapolated toward the ends...

The discrepancy between Carl Fogel's measurements and Frank Berto's
might be easy to explain. They measured different things. Carl
measured the rise of the tire with increasing pressure. I am not
surprised that this is not linear - the tire cannot expand
indefinitely.

Frank Berto measured how much pressure he needed for different
weights to achieve the same tire deflection of 15%. The 15% came from
tire manufacturers' recommendations. In Bicycle Quarterly's real road
tests, we found that performance decreased significantly once the
"tire drop" was more than about 13-15%. With less tire drop, higher
pressures did not yield significant performance benefits.

Please feel free to forward this clarification to rec.bike.tech.

Jan Heine
Editor
Bicycle Quarterly
140 Lakeside Ave #C
Seattle WA 98122
www.bikequarterly.com"

http://search.bikelist.org/getmsg.asp?Filename=internet-bob.10709.1869.eml

Dear Gary,

Anyone can load a tire, put a dial indicator on the top of the tire,
and measure the rise as inflation is increased in 10 psi increments.

It makes no difference whether the measurements are taken in
increasing or decreasing order.

The rise will _not_ be linear, with the effect particularly noticeable
at low pressures. If a straight line was extrapolated, that could
account for the discrepancy.

It would be fascinating to see the details of how the measurements
were achieved, since the tire deflects _most_ with the initial load.

In other words, the drop for 8 pounds of load is greatest for the
first 8 pounds (roughly the load of just a 20-lb bike on the front
tire). Measuring differences in thousandths of an inch gets awfully
tricky in practical terms out toward zero.

A 13%-15% absolute tire drop range is mentioned.

For a 1.000" tire, that's 0.130" to 0.150", a difference of 0.020".

On a ~1-inch tire, I found that a relative drop of 0.020" at ~90 psi
with an 85-lb load corresponded to a ~20 psi change in inflation.

Any bicyclist with a floor pump, a dial indicator, and some weights
can check my measurements.

You can pick up a dial indicator from Harbor Freight for about ten
bucks, less than the cost of a floor pump.

Cheers,

Carl Fogel

"Gary Young" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

On Fri, 28 Sep 2007 21:24:36 -0600, carlfogel wrote:

<snip>

Quoted message said:

.....In Bicycle Quarterly's real road tests, we found that
performance decreased significantly once the "tire drop" > was more than

about 13-15%. With less tire drop, higher > pressures did not yield
significant performance benefits.

Performance? Define performance. Was it an increase in rolling resistance?
More pedal force required? Or was it anecdotal subjective opinions of the
riders.

My personal opinion is that some riders like very hard tires and have
fooled themselves into thinking that tires run at high pressures will MAKE
them go faster.

If I'm going to ride on a rough surface road I ride Panaracer Paselas at
85-90PSI. If it's a smooth road surface I like Continental Grand Prixs at
95-110 PSI. I ride sewups at 90 PSI. That's MY preference based on ride
comfort and puncture potential.

Chas.

Quoted message said:

Some recent threads and an email today led me to measure how much a
fairly typical tire rose as I inflated it.

The results suggest that some people are fooling themselves.

First, some data from dial indicator measurements of a loaded tire's
height rise in thousandths of an inch from 40 to 130 psi, measured
from top of nominal 700x25c tire.

Tire measured 1.020" to 1.030" wide at top between 80 to 130 psi.

Total load on tire was ~85 pounds, resting on a scale, but the tire
stood on a concrete floor for the test.

inch mm inch
psi rise rise change
--- ----- ---- ------
40 0.000 0.00 n/a
50 0.021 0.53 0.021
60 0.038 0.97 0.017
70 0.052 1.32 0.014
80 0.063 1.60 0.011
90 0.076 1.93 0.013
100 0.088 2.24 0.012
110 0.100 2.54 0.012
120 0.110 2.79 0.010
130 0.120 3.05 0.010

My quick and dirty table shows only that the top of a roughly 1.020"
wide tire rose ~0.120 inches from 40 to 130 psi under an 88 pound
load.

For those interested in arbitrary figures, 15% of a tire width of
1.020" is 0.153", or 3.9 mm.

Presumably, someone could measure the diameter of a tire hanging in
mid-air and inflated to some arbitrary pressure, load it with some
weights, measure the change, and call it tire drop.

But this would require such an elaborate test rig that it would be
extremely difficult for any normal poster to even _try_ to make a
practical measurement of the difference between 90 psi and 100 psi.

After all, the difference between 90 and 100 psi with an elaborate
test rig using a dial indicator was only 0.012", about a hundredth of
an inch.

Incidentally, it is highly unlikely that pressure and a 700x25 tire's
rise have the straight linear relationship that some people suggest,
at least not over a 40 to 130 psi range with an 88 pound load.

I expect that beautifully smooth graphs of pressure and rise are just
as theoretical, oversimplified, and mistaken as similar predictions
for contact patch area, which does not follow the often-mentioned
tire-pressure/load = area prediction.

(Actual measurements of contact patches show that they fail to shrink
as much as predicted with higher pressures and fail to expand as much
as predicted with lower pressures. They stubbornly stick to a
preferred size.)

Cheers,

Carl Fogel

An email asked for details of the testing.

I wonder if other posters get as many emails as I do? Often it seems
as if people are accidentally hitting the email instead of the post
button on their newsreaders. I don't mind, but it would more fun if
they were all beautiful but lonely women.

Luckily, the test rig had been left untouched in hopes of gaining
Historic Place status, so here's a picture:

http://i24.tinypic.com/2iw4zn5.jpg

The white bathroom scales are not legal for trade, but showed ~85
pounds when shoved under the tire. They were removed for testing, lest
the tire inflation increased the contact patch area enough to cause
the scale surface to rise--remember, the measurements involved changes
as small as ten thousandths of an inch.

The yellow floor pump reads in alleged 2-lb increments. Zip ties
secure the pump head, which tended to come off during lengthy testing.
Fogel Labs has budgeted for a new O-ring.

The seat post sticks up between the jaws of the bench vise, which are
slightly open, but stop everything from toppling over. If the post had
been clamped, it might have held the front end of the bicycle down
slightly as the tire was pumped up--again, the measurements were so
tiny that such precautions were necessary.

The board was clamped to the red test frame for long-ago experiments.
Later, the weight bar was added for even stranger rituals.

The dial indicator from Sears is directly over the front axle and
contact patch. It rests on the weight bar, not the top of the tire,
partly because it was just about the right height, but mostly because
the slight expansion of the top of the tire during inflation would
have exaggerated the rise of the bottom of the tire.

A carefully machined indentation in the round weight bar assures that
the dial indicator does not slip off to the side and give false
readings. (It looks remarkably like a dent whacked with a blunt punch
and a hammer.)

The dial indicator is held by vise grips, which in turn are held in
place under a 10-pound weight. Vise grips were required because half
the convenient mounting lug on the back of the dial indicator was
hacksawed off in 1972 in order to fit into an awkward spot while
timing a motorcycle. I may spend ten bucks at Harbor Freight and get a
replacement, or maybe even a few bucks more for one with an adjustable
arm.

Cheers,

Carl Fogel

Tim McNamara said:

Heine measure loads on three bicycles with a rider seated on the bike
(randonneur bike, racing bike and "city" bike) and found that the weight
distribution (% front/rear) was 45/55, 30/60 and 35/65 respectively.

The middle figures don't add up. I'm guessing they should have been
40/60?

In article <[email hidden]>,

Quoted message said:

On Sat, 29 Sep 2007 14:23:23 -0500, Gary Young <[email hidden]>

Quoted message said:

carlfogel said:

On Fri, 28 Sep 2007 21:27:22 -0500, Tim McNamara
<[email hidden]> wrote:

>In article <[email hidden]>,
> [email hidden] wrote:

[long correspondence trimmed; interested viewers are invited to check
the archives]

Quoted message said:

Any bicyclist with a floor pump, a dial indicator, and some weights
can check my measurements.

You can pick up a dial indicator from Harbor Freight for about ten
bucks, less than the cost of a floor pump.

Aha! I am on to your craven attempts to encourage the readers of rbt to
engage in the oft-forsaken cornerstone of the Scientific Method:
reproducibility!

Well, we scholastics are on to you.

And it's a shame. I always assumed that a former literature explicator
like yourself would have had a naturally scholastic bent.

Silliness aside, I should take up your challenge, but my excuse is that
I have too many bikes a-building right now for others.

--
Ryan Cousineau [email hidden] http://www.wiredcola.com/
"I don't want kids who are thinking about going into mathematics
to think that they have to take drugs to succeed." -Paul Erdos

Quoted message said:

Some recent threads and an email today led me to measure how much a
fairly typical tire rose as I inflated it.

The results suggest that some people are fooling themselves.

First, some data from dial indicator measurements of a loaded tire's
height rise in thousandths of an inch from 40 to 130 psi, measured
from top of nominal 700x25c tire.

Tire measured 1.020" to 1.030" wide at top between 80 to 130 psi.

Total load on tire was ~85 pounds, resting on a scale, but the tire
stood on a concrete floor for the test.

inch mm inch
psi rise rise change
--- ----- ---- ------
40 0.000 0.00 n/a
50 0.021 0.53 0.021
60 0.038 0.97 0.017
70 0.052 1.32 0.014
80 0.063 1.60 0.011
90 0.076 1.93 0.013
100 0.088 2.24 0.012
110 0.100 2.54 0.012
120 0.110 2.79 0.010
130 0.120 3.05 0.010

My quick and dirty table shows only that the top of a roughly 1.020"
wide tire rose ~0.120 inches from 40 to 130 psi under an 88 pound
load.

For those interested in arbitrary figures, 15% of a tire width of
1.020" is 0.153", or 3.9 mm.

Presumably, someone could measure the diameter of a tire hanging in
mid-air and inflated to some arbitrary pressure, load it with some
weights, measure the change, and call it tire drop.

But this would require such an elaborate test rig that it would be
extremely difficult for any normal poster to even _try_ to make a
practical measurement of the difference between 90 psi and 100 psi.

After all, the difference between 90 and 100 psi with an elaborate
test rig using a dial indicator was only 0.012", about a hundredth of
an inch.

Incidentally, it is highly unlikely that pressure and a 700x25 tire's
rise have the straight linear relationship that some people suggest,
at least not over a 40 to 130 psi range with an 88 pound load.

I expect that beautifully smooth graphs of pressure and rise are just
as theoretical, oversimplified, and mistaken as similar predictions
for contact patch area, which does not follow the often-mentioned
tire-pressure/load = area prediction.

(Actual measurements of contact patches show that they fail to shrink
as much as predicted with higher pressures and fail to expand as much
as predicted with lower pressures. They stubbornly stick to a
preferred size.)

Cheers,

Carl Fogel

An odd example of measurement problems.

I took a few minutes and set up my dial calipers to hang and measure
the same tire's width as it sat in a truing stand. As I expected, the
tire widened as it was pumped up

psi width

30 0.975
40 0.985
50 0.992
60 0.996
70 0.998
80 1.007
90 1.014
100 1.018
110 1.025
120 1.029
130 1.034
140 1.040

But trying to get a good reading by rolling a dial caliper's jaws shut
on squishy rubber _felt_ a little squishy and inaccurate, even though
the results looked good.

Rather than adjusting the dial caliper to slip at less and less
resistance as its jaws closed on the rubber, I decided to try a dial
indicator, which supplies its own very gentle spring action.

Naturally, I got sidetracked and first set the dial indicator up to
measure the expansion of the top of the tire in the truing stand, not
its sideways bulge. But things turned out pretty well:

top of
psi tire

? 1.000 as low as I could get it and still feel pressure
10 0.990 probably seated a bit, moving the wrong way
20 0.994
30 0.998
40 1.002
50 1.007
60 1.011
70 1.016
80 1.023
90 1.028
100 1.033
110 1.038
120 1.042
130 1.050
140 pump head blew off

Then I set things up again in the truing stand, but turned the dial
indicator rig on the side of the tire to measure how far one side of
the tire bulged sideways as I pumped it up.

I zeroed the dial indicator, checked that everything was at right
angles, and pumped . . . Nothing.

No change at 10 psi.

I pumped to 20 psi, but still nothing happened.

At 30 psi, the needle still sat on 0.000.

I went up to 80 psi without any movement.

At 120 psi, there was a little movement, about 0.003 inches.

Mystified, I grabbed my dial calipers, closed their jaws on the stupid
tire, and let the air out of the dumb thing. A gratifying gap promptly
opened between the side of the tire and the jaws, so I knew that I
wasn't crazy--tires really do expand when inflated.

But another attempt with the dial _indicator_ produced the same
failure.

At first, I thought that the problem was just that the tire was
expanding outward, as well as sideways, and the dial indicator was
slipping down the side of the expanding tire, just enough for the loss
to cancel the gain.

Some of the failure was due to this movement in two directions, but I
suspect that real problem was the tiny bit of stiction in the dial
indicator rod as the rubber moved outward as well as sideways may have
made things worse, too, jamming the delicate mechanism.

Whatever the cause, the dial indicator utterly failed to measure the
sideways expansion clearly indicated by dial calipers.

Cheers,

Carl Fogel

On Sun, 30 Sep 2007 05:10:12 GMT, Ryan Cousineau <[email hidden]>
wrote:

[snip]

Quoted message said:

Silliness aside, I should take up your challenge, but my excuse is that
I have too many bikes a-building right now for others.

Dear Ryan,

Just remember that it could be a very bad thing if a bike with 80
pounds of weights at handlebar height topples over onto you.

Safety rope the rig to a work bench or arrange for a hefty vise's jaws
to cuddle a stout part of the bike frame.

But do _something_ to make sure that your carefully balanced test bike
rig can't fall over and hurt you.

I haven't had any accidents so far, and I don't want to read about
anyone else having different results.

Er, I mean anyone else having accidents. Different results are fine.

Cheers,

Carl Fogel