Frame flex and efficiency



Tom Ace writes:


>>> 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.


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


> 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
 
[email protected] wrote:

>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.


>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.


>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.


>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 writes:

>> 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.


>> 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.


>> 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.


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

>> 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".


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.

> 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.

> 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 wrote:

> 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?


> 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
 
On Mon, 08 Aug 2005 23:33:40 GMT,
[email protected] wrote:

>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
 
[email protected] wrote:
>
> 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 protected]> wrote in message
news:[email protected]...
> [email protected] wrote:
>
>>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.
>
>
>>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.
>
>
>>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.
>
>
>>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
 
On 8 Aug 2005 18:01:24 -0700, "Ron Ruff"
<[email protected]> wrote:

>[email protected] wrote:
>>
>> 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 writes:

>>> 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.


> 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.


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

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

> 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.

> 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.

> (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
 
[email protected] wrote:

> 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.

> 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.

> > 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
 
[email protected] wrote:

> 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.

> 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.

> > 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" <[email protected]> wrote:

>[email protected] wrote:
>
>> > 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 protected]
Madison, WI 53711 USA
-----------------------------------------------
 
On Mon, 08 Aug 2005 23:13:14 -0500, Jim Adney
<[email protected]> wrote:

>On 7 Aug 2005 21:43:41 -0700 "Tom Ace" <[email protected]> wrote:
>
>>[email protected] wrote:
>>
>>> > 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 protected]
> Madison, WI 53711 USA
>-----------------------------------------------


Dear Jim,

[I asked}

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


[Tom Ace replied]

>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 wrote:

> 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 wrote:

> 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.

> 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".

> > 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 wrote:
> 41 wrote:
>
> > 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 protected]> wrote in message
news:[email protected]...
> Philip Holman wrote:
>
>> 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.

> 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 wrote:
> Tom Ace wrote:
>
>>41 wrote:
>>
>>
>>>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 wrote:
> 41 wrote:


> > 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.

> 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.