Frame flex and efficiency

Tom Ace said:

Quoted message said:

Quoted message said:

In the stationary-bike demonstration, I can lower the outside of
the left pedal by about one inch. This is with a Columbus SL
frame (from the late 1970s) and 170mm cranks, and I weigh 85kg.

Quoted message said:

Quoted message said:

Does this drop take into account the flattening of the tires with
the load?

Quoted message said:

The one inch doesn't include the tire squashage. The pedal drops by
a little over an inch relative to the floor.

OK, so at what pedal position are you doing this and how. Are you
sitting astride the bicycle in the saddle? How is this bicycle
anchored to the trainer? This sounds like a huge displacement for a
bicycle on which a rider is pedaling.

Jobst Brandt

Quoted message said:

OK, so at what pedal position are you doing this and how.

The bike is on a hard floor. The cranks are horizontal.
I use the rear brake to keep the rear wheel from turning.
A marker mounted on the pedal draws on a stationary piece
of paper.

Quoted message said:

Are you sitting astride the bicycle in the saddle?

Did you read the part where I'd said "stand alongside the bike"?
My left foot is on the floor, my right foot is on the left pedal.

Quoted message said:

How is this bicycle anchored to the trainer?

There is no trainer. My eye is over the frame and I visually
ensure that the bike stays upright. I am satisfied I can do this
well enough to keep the error due to lean less than +/- 0.2 inch.

Quoted message said:

This sounds like a huge displacement for a bicycle on which
a rider is pedaling.

But I didn't make a claim about normal riding and pedaling.
I said the pedal moves noticeably in a demonstration with a
stationary bike; you said "How much? How about measuring it.
I think you'll fins [sic] it is tiny." My procedure is good
enough to show that the pedal moves more than I think you
had in mind when you said "tiny".

I'd expect that while riding uphill (and while able to pull
up on the other pedal) I could even apply more force than I
did in a stationary-bike demo.

I find it curious that you asked me to make a measurement
and then asked for the details of my procedure, while you
feel free to make seat-of-the-pants claims like "My bicycle
does not produce such deflections when I am climbing."

Tom Ace

Tom Ace said:

Quoted message said:

OK, so at what pedal position are you doing this and how.

Quoted message said:

The bike is on a hard floor. The cranks are horizontal. I use the
rear brake to keep the rear wheel from turning. A marker mounted on
the pedal draws on a stationary piece of paper.

Quoted message said:

Quoted message said:

Are you sitting astride the bicycle in the saddle?

Quoted message said:

Did you read the part where I'd said "stand alongside the bike"? My
left foot is on the floor, my right foot is on the left pedal.

Quoted message said:

Quoted message said:

How is this bicycle anchored to the trainer?

Quoted message said:

There is no trainer. My eye is over the frame and I visually ensure
that the bike stays upright. I am satisfied I can do this well
enough to keep the error due to lean less than +/- 0.2 inch.

When you said "stationary bicycle" I assumed you meant trainer.

Quoted message said:

Quoted message said:

This sounds like a huge displacement for a bicycle on which a rider
is pedaling.

Quoted message said:

But I didn't make a claim about normal riding and pedaling. I said
the pedal moves noticeably in a demonstration with a stationary
bike; you said "How much? How about measuring it. I think you'll
fins [sic] it is tiny." My procedure is good enough to show that
the pedal moves more than I think you had in mind when you said
"tiny".

This is much like the common lateral frame flex test done by pushing
with one foot from the side. I don't doubt that a bicycle frame is
elastic but I don't believe the pedal crank can be made to rotate
around its BB by one inch when pressing down on the pedal. That is
what your description seemed to imply.

Quoted message said:

I'd expect that while riding uphill (and while able to pull up on
the other pedal) I could even apply more force than I did in a
stationary-bike demo.

I think you can do that statically by leaning your shoulder against a
wall while locking the rear brake. I am curious where you believe the
rotational flex is being generated. I don't see where.

Quoted message said:

I find it curious that you asked me to make a measurement and then
asked for the details of my procedure, while you feel free to make
seat-of-the-pants claims like "My bicycle does not produce such
deflections when I am climbing."

Why is that curious when I can see practically no motion when I do
such a test on my bicycle with the crank horizontal. You must be
doing something differently. As you said, you were off the bicycle
pushing from one side with the "other" foot. I don't think that is a
valid simulation of frame flex when pedaling.

Jobst Brandt

Tom Ace said:

The bike is on a hard floor. The cranks are horizontal.
I use the rear brake to keep the rear wheel from turning.
A marker m ounted on the pedal draws on a stationary piece
of paper.

Try it without tires mounted

How exactly is the marker mounted? Does it amplify the motion?

Quoted message said:

There is no trainer. My eye is over the frame and I visually
ensure that the bike stays upright. I am satisfied I can do this
well enough to keep the error due to lean less than +/- 0.2 i nch.

How did you satisfy yourself of this?

Apropos percent grade.

The reason we express gradients in "percent" (%)or for that matter in
"permil" (0/00) for railroads, is that the weight of the vehicle times
the grade gives the force required to move forward against that
gradient purely from slope considerations. That way we don't have to
get into trigonometry and other indirect means to assess how much
force/power is required to move forward.

Jobst Brandt

Quoted message said:

Apropos percent grade.

The reason we express gradients in "percent" (%)or for that matter in
"permil" (0/00) for railroads, is that the weight of the vehicle times
the grade gives the force required to move forward against that
gradient purely from slope considerations. That way we don't have to
get into trigonometry and other indirect means to assess how much
force/power is required to move forward.

Jobst Brandt

Dear Jobst,

Multiplying the weight by the grade% works for gentle
railroad grades, but over-predicts more and more for steeper
grades.

Here's a quick table for a 200 lb weight on various grades
to illustrate how the grade% method predicts a larger force
for steeper angles than the sine method:

grade predicted angle of sine of predicted
% slope grade% angle slope
force force
5% 5.00 2.862 0.4999 4.99 steep rr
10% 10.00 5.711 0.0995 9.95
15% 15.00 8.531 0.1483 14.83
20% 20.00 11.310 0.1961 19.61
25% 25.00 14.036 0.2425 24.25
30% 30.00 16.699 0.2873 28.73 fargo st.
35% 35.00 19.290 0.3304 33.04
40% 40.00 21.801 0.3714 37.14
45% 45.00 24.228 0.4104 41.04
50% 50.00 26.565 0.4472 44.72 10% off
. . .
100% 100.00 45.000 0.7071 70.71 41% off

This uses the grade% = tan(angle) method. (Grade% can be
calculated as rise/run, but run can be either the shorter
theoretical horizontal adjacent side of the right triangle,
or the slightly longer hypotenuse that is the actual
distance that we pedal.)

Carl Fogel

Quoted message said:

Multiplying the weight by the grade% works for gentle
railroad grades, but over-predicts more and more for steeper
grades.

This uses the grade% = tan(angle) method. (Grade% can be
calculated as rise/run, but run can be either the shorter
theoretical horizontal adjacent side of the right triangle,
or the slightly longer hypotenuse that is the actual
distance that we pedal.)

Grade is the rise in elevation divided by the "actual distance we
pedal" (the hypotenuse)... not the horizontal distance. That is why you
are getting an error.

-Ron

"Tom Ace" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

Quoted message said:

OK, so at what pedal position are you doing this and how.

The bike is on a hard floor. The cranks are horizontal.
I use the rear brake to keep the rear wheel from turning.
A marker mounted on the pedal draws on a stationary piece
of paper.

Quoted message said:

Are you sitting astride the bicycle in the saddle?

Did you read the part where I'd said "stand alongside the bike"?
My left foot is on the floor, my right foot is on the left pedal.

Quoted message said:

How is this bicycle anchored to the trainer?

There is no trainer. My eye is over the frame and I visually
ensure that the bike stays upright. I am satisfied I can do this
well enough to keep the error due to lean less than +/- 0.2 inch.

Quoted message said:

This sounds like a huge displacement for a bicycle on which
a rider is pedaling.

But I didn't make a claim about normal riding and pedaling.
I said the pedal moves noticeably in a demonstration with a
stationary bike; you said "How much? How about measuring it.
I think you'll fins [sic] it is tiny." My procedure is good
enough to show that the pedal moves more than I think you
had in mind when you said "tiny".

I'd expect that while riding uphill (and while able to pull
up on the other pedal) I could even apply more force than I
did in a stationary-bike demo.

I find it curious that you asked me to make a measurement
and then asked for the details of my procedure, while you
feel free to make seat-of-the-pants claims like "My bicycle
does not produce such deflections when I am climbing."

Not only did you measure frame flex but also the delection of the whole
drive train from pedal to brake attactment. Try it again and compare the
difference between the extremes of your gears. From the largest
chainring/smallest sproket to the smallest chainring/largest sproket
combos. Please let us know the result.

Phil H

Ron Ruff said:

Quoted message said:

Multiplying the weight by the grade% works for gentle
railroad grades, but over-predicts more and more for steeper
grades.

This uses the grade% = tan(angle) method. (Grade% can be
calculated as rise/run, but run can be either the shorter
theoretical horizontal adjacent side of the right triangle,
or the slightly longer hypotenuse that is the actual
distance that we pedal.)

Grade is the rise in elevation divided by the "actual distance we
pedal" (the hypotenuse)... not the horizontal distance. That is why you
are getting an error.

-Ron

Dear Ron,

Both methods are used, since at railroad grades the
difference is so tiny that it's lost in the normal error of
construction.

I think that we agree about the reason for the error.

A 100% grade is usually considered 45 degrees with no error
in converting degrees to grade, even though it's not 1/1
(100%), but rather 1/1.41 (71%) by the rise/travel method
reflected by geometry and the tangent.

(Or so I think--maybe someone has a link to a table in which
45 degrees is considered 71% grade. The tables that actually
care about such methods are usually limited to railroad
grades well under 10%.)

But in physics, as I understand it, the slope force is
calculated by multiplying the weight on the slope by the
sine of the angle.

The grade x weight method predicts a 100 lb slope force for
a 100 lb weight on a 100% grade 45-degree slope.

The sine x weight method predicts a 70.71 slope force for
the same 100 lb weight on a 100% 45-degree slope.

I actually tried both ways when I was trying to figure out
why Jobst's 326 pound figure was so far from what I was
getting on a 31.5% grade.

(The explanation there may be that Jobst was trying to
illustrate high pedal force and remembering a calculation
involving a 17-tooth third gear, not the 20 or 21 tooth that
he mentioned.)

Carl Fogel

Carl Fogel said:

Quoted message said:

Quoted message said:

Multiplying the weight by the grade% works for gentle railroad
grades, but over-predicts more and more for steeper grades.

Quoted message said:

Quoted message said:

Quoted message said:

This uses the grade% = tan(angle) method. (Grade% can be
calculated as rise/run, but run can be either the shorter
theoretical horizontal adjacent side of the right triangle, or the
slightly longer hypotenuse that is the actual distance that we
pedal.)

Quoted message said:

Quoted message said:

Grade is the rise in elevation divided by the "actual distance we
pedal" (the hypotenuse)... not the horizontal distance. That is why
you are getting an error.

Quoted message said:

Both methods are used, since at railroad grades the
difference is so tiny that it's lost in the normal error of
construction.

Quoted message said:

I think that we agree about the reason for the error.

Quoted message said:

A 100% grade is usually considered 45 degrees with no error in
converting degrees to grade, even though it's not 1/1 (100%), but
rather 1/1.41 (71%) by the rise/travel method reflected by geometry
and the tangent.

In road and RR design, gradient IS the tangent and 100% grade IS 45
degrees.

Quoted message said:

(Or so I think--maybe someone has a link to a table in which
45 degrees is considered 71% grade. The tables that actually
care about such methods are usually limited to railroad
grades well under 10%.)

Quoted message said:

But in physics, as I understand it, the slope force is calculated by
multiplying the weight on the slope by the sine of the angle.

Quoted message said:

The grade x weight method predicts a 100 lb slope force for
a 100 lb weight on a 100% grade 45-degree slope.

The work done is grade x weight x horizontal distance... which amounts
to how much elevation is gained per unit movement horizontally.

Quoted message said:

The sine x weight method predicts a 70.71 slope force for the same
100 lb weight on a 100% 45-degree slope.

That's because you are traveling farther under a proportionately
lesser force by that method.

Quoted message said:

I actually tried both ways when I was trying to figure out why
Jobst's 326 pound figure was so far from what I was getting on a
31.5% grade.

Fuzzy memory and thinking.

Quoted message said:

(The explanation there may be that Jobst was trying to illustrate
high pedal force and remembering a calculation involving a 17-tooth
third gear, not the 20 or 21 tooth that he mentioned.)

Not so. I was using a 47-50 CW ND 13,15,17,20,25t cluster. It was
later that I switched to 46-50 CW and 13,15,17,19,21,24t cluster.

Jobst Brandt

Quoted message said:

This is much like the common lateral frame flex test done by pushing
with one foot from the side. I don't doubt that a bicycle frame is
elastic but I don't believe the pedal crank can be made to rotate
around its BB by one inch when pressing down on the pedal. That is
what your description seemed to imply.

My description was about a static test. The measurement is
consistent and repeatable. If you don't believe me, come
to Lone Pine some time and I'll show you.

Quoted message said:

I think you can do that statically by leaning your shoulder against a
wall while locking the rear brake. I am curious where you believe the
rotational flex is being generated. I don't see where.

I wasn't making claims about exactly what flexed;
I was reporting how far the pedal moved in my experiment.

Quoted message said:

Quoted message said:

I find it curious that you asked me to make a measurement and then
asked for the details of my procedure, while you feel free to make
seat-of-the-pants claims like "My bicycle does not produce such
deflections when I am climbing."

Why is that curious when I can see practically no motion when I do
such a test on my bicycle with the crank horizontal. You must be
doing something differently. As you said, you were off the bicycle
pushing from one side with the "other" foot. I don't think that is a
valid simulation of frame flex when pedaling.

The stand-to-the-side test is more artificial, yes.

After reading your response this time, I tested again as you
described: sitting on the saddle, leaning against something to
keep the bike vertical. Subjectively--that is, before I did
the marker-on-paper test--I had the impression that the pedal
wasn't moving as far. It looks more dramatic when you are
standing to the side. But when I measured, I still saw the
pedal move one inch. This is without pulling up on the other pedal.

Because you've described the effect in word ("tiny"😉 rather
than number, I get the impression you haven't measured it.

Tom Ace

Quoted message said:

This is much like the common lateral frame flex test done by pushing
with one foot from the side. I don't doubt that a bicycle frame is
elastic but I don't believe the pedal crank can be made to rotate
around its BB by one inch when pressing down on the pedal. That is
what your description seemed to imply.

My description was about a static test. The measurement is
consistent and repeatable. If you don't believe me, come
to Lone Pine some time and I'll show you.

Quoted message said:

I think you can do that statically by leaning your shoulder against a
wall while locking the rear brake. I am curious where you believe the
rotational flex is being generated. I don't see where.

I wasn't making claims about exactly what flexed;
I was reporting how far the pedal moved in my experiment.

Quoted message said:

Quoted message said:

I find it curious that you asked me to make a measurement and then
asked for the details of my procedure, while you feel free to make
seat-of-the-pants claims like "My bicycle does not produce such
deflections when I am climbing."

Why is that curious when I can see practically no motion when I do
such a test on my bicycle with the crank horizontal. You must be
doing something differently. As you said, you were off the bicycle
pushing from one side with the "other" foot. I don't think that is a
valid simulation of frame flex when pedaling.

The stand-to-the-side test is more artificial, yes.

After reading your response this time, I tested again as you
described: sitting on the saddle, leaning against something to
keep the bike vertical. Subjectively--that is, before I did
the marker-on-paper test--I had the impression that the pedal
wasn't moving as far. It looks more dramatic when you are
standing to the side. But when I measured, I still saw the
pedal move one inch. This is without pulling up on the other pedal.

Because you've described the effect in word ("tiny"😉 rather
than number, I get the impression you haven't measured it.

Tom Ace

On 7 Aug 2005 21:43:41 -0700 'Tom Ace said:

Quoted message said:

Quoted message said:

In the demonstration I described, the pedal gets closer to the
ground (most noticeably at its outermost point). It's not like BB
displacement is the only thing going on; there's torsion as well.

How much? How about measuring it. I think you'll find it is tiny.
Of course you need to restrain the bicycle so you are measuring the
right thing and not just leaning the bicycle.

In the stationary-bike demonstration, I can lower the outside
of the left pedal by about one inch. This is with a Columbus SL
frame (from the late 1970s) and 170mm cranks, and I weigh 85kg.

Sorry, I don't have your previous post, so this comment may be moot,
but how much of this motion is in compression of the tires? If you're
just standing next to the bike and pressing on one pedal, then this is
different from the case where you're already on the bike and the tires
are already loaded and compressed.

-
-----------------------------------------------
Jim Adney [email hidden]
Madison, WI 53711 USA
-----------------------------------------------

Jim Adney said:

On 7 Aug 2005 21:43:41 -0700 'Tom Ace said:

Quoted message said:

> In the demonstration I described, the pedal gets closer to the
> ground (most noticeably at its outermost point). It's not like BB
> displacement is the only thing going on; there's torsion as well.

How much? How about measuring it. I think you'll find it is tiny.
Of course you need to restrain the bicycle so you are measuring the
right thing and not just leaning the bicycle.

In the stationary-bike demonstration, I can lower the outside
of the left pedal by about one inch. This is with a Columbus SL
frame (from the late 1970s) and 170mm cranks, and I weigh 85kg.

Sorry, I don't have your previous post, so this comment may be moot,
but how much of this motion is in compression of the tires? If you're
just standing next to the bike and pressing on one pedal, then this is
different from the case where you're already on the bike and the tires
are already loaded and compressed.

-
-----------------------------------------------
Jim Adney [email hidden]
Madison, WI 53711 USA
-----------------------------------------------

Dear Jim,

[I asked}

Quoted message said:

Quoted message said:

Does this drop take into account the flattening of the tires
with the load?

[Tom Ace replied]

Quoted message said:

The one inch doesn't include the tire squashage.
The pedal drops by a little over an inch relative to the floor.

I couldn't resist repeating "squashage"--some coinings are
too good to ignore.

Carl Fogel

Philip Holman said:

Not only did you measure frame flex but also the delection of the whole
drive train from pedal to brake attactment. Try it again and compare the
difference between the extremes of your gears. From the largest
chainring/smallest sproket to the smallest chainring/largest sproket
combos. Please let us know the result.

In high gear, with both brakes on, and with the front
tire against a wall, I doubt that brake flex is
contributing much to the measurement. With those
conditions (47/13 gear, 23-622 tires) I see the pedal
move one inch. In low gear (41/34) it moves about 1.25".

This experiment is generating more interest than I'd expected.
You can try it yourself, y'know. It won't take long, and you
don't need an assistant.

Tom Ace

41 said:

Try it without tires mounted

I love these commands. At least Philip Holman said "please".

I did it again with me on the bike (i.e., without a foot on
the floor). That way, the tires bear my weight at all times.

Quoted message said:

How exactly is the marker mounted? Does it amplify the motion?

The marker is held against the underside of the pedal with a
spring-loaded woodworking clamp. It's not wiggling during
the test. It does stick out a little past the end of the
pedal, so amplification is possible if flex causes the pedal
spindle to not be parallel to the ground, but that's a small
effect. I wasn't going for the ultimate in precision, just
to refute Jobst's assertion that the effect is "tiny".

Quoted message said:

Quoted message said:

There is no trainer. My eye is over the frame and I visually
ensure that the bike stays upright. I am satisfied I can do this
well enough to keep the error due to lean less than +/- 0.2 i nch.

How did you satisfy yourself of this?

I observed how much lean was required to drop the pedal
by 0.2 inch. It's not hard to keep the bike from leaning
considerably less than that; I was being conservative with
the error limit I gave.

Tom Ace

Tom Ace said:

41 said:

Try it without tires mounted

I love these commands. At least Philip Holman said "please".

Command mode is "Do it without tires mounted!" "Try" is usually
considered fairly close to "please"- so much so that you didn't seem to
notice that PH in fact did the same thing. He only asked please for the
report...

Please explain the following!

Roughly speaking: for a 60cm frame, if the line connecting the seat lug
to pedal/marker is at about a 20 degree angle from the plumb, and if
the motion were such that the seat lug remained stationary, a 1 inch
drop would corresond to about a 7cm or nearly 3 inch lateral
translation of the BB. In bicycling magazine tests they get it to move
2-5 mm when they simulate such forces in a fixture.

"Tom Ace" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

Philip Holman said:

Not only did you measure frame flex but also the delection of the
whole
drive train from pedal to brake attactment. Try it again and compare
the
difference between the extremes of your gears. From the largest
chainring/smallest sproket to the smallest chainring/largest sproket
combos. Please let us know the result.

In high gear, with both brakes on, and with the front
tire against a wall, I doubt that brake flex is
contributing much to the measurement. With those
conditions (47/13 gear, 23-622 tires) I see the pedal
move one inch. In low gear (41/34) it moves about 1.25".

Which would be ~ .875" with zero torque on the back wheel (imagine a
seized BB). Quite believeable.

Quoted message said:

This experiment is generating more interest than I'd expected.
You can try it yourself, y'know. It won't take long, and you
don't need an assistant.

You are doing just fine with your measurements.

We didn't get to see any direct overhead shots of the TdF sprint
finishes this year; those fish-tail bike deflections are large enough to
see.

Phil H

41 said:

Tom Ace said:

41 said:

Try it without tires mounted

I love these commands. At least Philip Holman said "please".

Command mode is "Do it without tires mounted!" "Try" is usually
considered fairly close to "please"- so much so that you didn't seem to
notice that PH in fact did the same thing. He only asked please for the
report...

Please explain the following!

Roughly speaking: for a 60cm frame, if the line connecting the seat lug
to pedal/marker is at about a 20 degree angle from the plumb, and if
the motion were such that the seat lug remained stationary, a 1 inch
drop would corresond to about a 7cm or nearly 3 inch lateral
translation of the BB. In bicycling magazine tests they get it to move
2-5 mm when they simulate such forces in a fixture.

Your calculation completely ignores twisting or torsion of the
bottom bracket relative to the average plane of the frame. This
is very significant, making your required 3" translation nonsense.

Tom has made actual measurements, not speculation based on
assumptions of how a frame reacts under load.

Dave Lehnen

Dave Lehnen said:

41 wrote:

Quoted message said:

Quoted message said:

Roughly speaking: for a 60cm fram e, if the line connecting the seat lug
to pedal/marker is at about a 20 degree angle from the plumb, and if
the motion were such that the seat lug remained stationary, a 1 inch
drop would corresond to about a 7cm or nearly 3 inch lateral
t ranslation of the BB. In bicycling magazine tests they get it to move
2-5 mm when they simulate such forces in a fixture.

Your calculation completely ignores twisting or torsion of the
bottom bracket relative to the average plane of the fr ame.

Uh, no it doesn't, even though your explanation is not entirely clear.
As I said this is a rough calculation but it is what it is represented
as. Give a calculation of your own of the translation if you don't like
mine.

Quoted message said:

Tom has made actual measurements, not speculation based on
assumptions of how a frame reacts under load.

Actual measurements of what is the question, and this requires some
thinking: think twice, measure once. He has done thinking but perhaps
more is useful. For example, his method ignores flex of BB spindle, hub
windup, brake flex / front tire compression at horizontal, and other
such things. That he could get a quarter inch difference just by
changing gears show that these have a real effect. He tried with pedal
at horizontal, it would have been better with pedal vertical (down). I
can't compare if only because I weigh so much less and my frame is
heavier gauge.

And as I said bike magazines which test for the same in jigs get only a
few mm deviation at the BB.