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Fat tire riders look like "fat heads."



 
 
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  #51  
Old July 7th 20, 05:19 PM posted to rec.bicycles.tech
Frank Krygowski[_4_]
external usenet poster
 
Posts: 10,538
Default Fat tire riders look like "fat heads."

On 7/7/2020 1:23 AM, John B. wrote:
On Tue, 7 Jul 2020 03:56:54 +0000 (UTC), Ralph Barone
wrote:

John B. wrote:
On Tue, 7 Jul 2020 00:52:05 +0000 (UTC), Ralph Barone
wrote:

John B. wrote:
On Mon, 6 Jul 2020 10:41:58 -0700 (PDT), Lou Holtman
wrote:

On Monday, July 6, 2020 at 7:15:32 PM UTC+2, Mark J. wrote:



It's a topic for another post, but I find on the road that energy /per
mile/ is surprisingly consistent, ranging from around 27 kJ/mile for a
gentle pootle to maybe 35 kJ/mile for a very brisk hilly ride, with most
rides in the 30-33 range. Not, of course, on rides that start and end
at different elevations.


Tell that to the people that say that 10-20 Watt on average is insignificant.

Lou

Well, it is the difference between 0.01 and 0.02 H.P. :-)
--
Cheers,

John B.



If 1000 W = 1 HP, yes...

I believe that the usual calculation is 1 hp = 745.69987 watts.
--
Cheers,

John B.



That’s what I thought too, but 1 HP = 1000 W was the only way to explain
your previous post. Maybe you’ve got better horses over there :-)



Well granted I did round a fairly lengthy number but no, 1 hp /
745.69987 watts = 0.0013410220896 hp.
So 10 watts is actual 0.013410220896 hp and 20 watts = 0.0268204417919
hp



Are you SURE about that last significant figure? After all, this is
regarding bicycling, where every minute change is critical! ;-)


--
- Frank Krygowski
Ads
  #52  
Old July 7th 20, 05:25 PM posted to rec.bicycles.tech
Frank Krygowski[_4_]
external usenet poster
 
Posts: 10,538
Default Fat tire riders look like "fat heads."

On 7/7/2020 12:14 AM, Mark J. wrote:
On 7/6/2020 5:40 PM, Frank Krygowski wrote:
On 7/6/2020 1:15 PM, Mark J. wrote:


There's no doubt I'm sinking a lot more energy into rolling
resistance than on a road bike.* I can even quantify it a bit, being
a data junkie and having PowerTap wheels on both road and gravel
bikes.* On the road, it takes a very brisk ride to burn over about
33-34 kiloJoules per mile. **Yesterday on the gravel, at a much lower
speed, I averaged 35 kJ/mile. **This is for rides that start and end
at the same elevation.

It's a topic for another post, but I find on the road that energy
/per mile/ is surprisingly consistent, ranging from around 27 kJ/mile
for a gentle pootle to maybe 35 kJ/mile for a very brisk hilly ride,
with most rides in the 30-33 range.* Not, of course, on rides that
start and end at different elevations.


I don't recall coming across kJ/mile numbers before, but looking at
the units, they reduce to units of force. So what those numbers
represent is the average force applied to drive the bike forward over
the course of the ride. So that would equal the total drag force on
the bike, more or less - rolling resistance, air resistance, what's
lost in jiggling the rider's body, etc.

Working the conversions, I get 30 kJ/mile = 18.6 Newtons force, or 4.2
pounds drag force on the bike, on average.

Air drag varies as the square of speed; but I'd expect for a
reasonable variation in low-ish speeds, the air drag would be at least
fairly constant. I'm assuming you're not riding over 20 mph.


Generally not, unless downhill/downwind.

That's an interesting way to think about it.* I've mostly been crudely
equating kJ/mile with Cal/mile - that's Cal *input*, *not* output -
since ~25% biomechanical efficiency of cycling means 1 Cal input ~= 1 kJ
output.* [1 Cal = 1 kcal = 4.184 kJ].* So I've been taking a very
different point of view.

Several years back I was teaching a freshman course that included
dimensional analysis and unit conversion.* Being only a little creative
you can "convert" cookies eaten into miles cycled.* I showed 'em that
one big cookie will get you 10 miles down the road.* A student replied
"no way that's right."* Cycling is vastly more efficient than most
people realize.


Right, cycling is amazingly efficient. That's one of its many beauties.

It also explains the "aerobellies" that are common on so many older club
cyclists, in my experience. (Although, thankfully, not my _personal_
experieence.) I think people give themselves too much credit for their
riding exercise, and overcompensate when eating.

On the other hand: One friend of mine retired just a couple years ago.
He's taken on riding as an almost-full-time job, and is well on track to
break 10,000 miles this year. In two years, he's lost something like 40
pounds!

Maybe he just doesn't have time to eat?


--
- Frank Krygowski
  #53  
Old July 7th 20, 09:04 PM posted to rec.bicycles.tech
Ralph Barone[_4_]
external usenet poster
 
Posts: 853
Default Fat tire riders look like "fat heads."

Frank Krygowski wrote:
On 7/6/2020 10:19 PM, Ralph Barone wrote:
Jeff Liebermann wrote:
On Tue, 07 Jul 2020 07:20:22 +0700, John B.
wrote:

Doesn't wind resistance increase as a square of the velocity?

Yep. Wind resistance is drag and is proportional to the square of the
velocity. However, the energy needed to overcome that drag increases
with the cube of the velocity:
https://physics.info/drag/
See the section on "drag and power":
...if drag is proportional to the square of speed,
then the power needed to overcome that drag is proportional
to the cube of speed (P ? v3). You want to ride your
bicycle twice as fast, you'll have to be eight times
more powerful. This is why motorcycles are so much
faster than bicycles.



Right, but you get there in half the time, so the energy required to go a
certain distance should go up as the square of the speed, once you reach a
velocity where aerodynamic drag dominates.


I'd leave Time out of it.

I'd just say that Energy, like Work, is Force times Distance. If you
were to double your speed, the Force required would indeed increase by a
factor of 4 (i.e. the square of speed, as you said). Multiplying that by
the same distance traveled yields four times the Work (or Energy) required.

I'm sure there are other ways of thinking about this, but my explanation
seems simplest to me.



Yes, that’s the “clean sheet of paper” approach, but I was trying to get
from Jeff’s post back to Mark’s conclusion.

  #54  
Old July 8th 20, 01:46 AM posted to rec.bicycles.tech
John B.[_3_]
external usenet poster
 
Posts: 5,697
Default Fat tire riders look like "fat heads."

On Tue, 7 Jul 2020 12:19:55 -0400, Frank Krygowski
wrote:

On 7/7/2020 1:23 AM, John B. wrote:
On Tue, 7 Jul 2020 03:56:54 +0000 (UTC), Ralph Barone
wrote:

John B. wrote:
On Tue, 7 Jul 2020 00:52:05 +0000 (UTC), Ralph Barone
wrote:

John B. wrote:
On Mon, 6 Jul 2020 10:41:58 -0700 (PDT), Lou Holtman
wrote:

On Monday, July 6, 2020 at 7:15:32 PM UTC+2, Mark J. wrote:



It's a topic for another post, but I find on the road that energy /per
mile/ is surprisingly consistent, ranging from around 27 kJ/mile for a
gentle pootle to maybe 35 kJ/mile for a very brisk hilly ride, with most
rides in the 30-33 range. Not, of course, on rides that start and end
at different elevations.


Tell that to the people that say that 10-20 Watt on average is insignificant.

Lou

Well, it is the difference between 0.01 and 0.02 H.P. :-)
--
Cheers,

John B.



If 1000 W = 1 HP, yes...

I believe that the usual calculation is 1 hp = 745.69987 watts.
--
Cheers,

John B.



Thats what I thought too, but 1 HP = 1000 W was the only way to explain
your previous post. Maybe youve got better horses over there :-)



Well granted I did round a fairly lengthy number but no, 1 hp /
745.69987 watts = 0.0013410220896 hp.
So 10 watts is actual 0.013410220896 hp and 20 watts = 0.0268204417919
hp



Are you SURE about that last significant figure? After all, this is
regarding bicycling, where every minute change is critical! ;-)


Ah well, my calculator were "rounded" a bit so perhaps my numbers are
a bit off. So in the pursuit of greater accuracy one watt should be
referenced as 0.00134102209244049000 hp.

--
Cheers,

John B.

  #55  
Old July 8th 20, 02:33 AM posted to rec.bicycles.tech
Frank Krygowski[_4_]
external usenet poster
 
Posts: 10,538
Default Fat tire riders look like "fat heads."

On 7/7/2020 8:46 PM, John B. wrote:
On Tue, 7 Jul 2020 12:19:55 -0400, Frank Krygowski
wrote:

On 7/7/2020 1:23 AM, John B. wrote:
On Tue, 7 Jul 2020 03:56:54 +0000 (UTC), Ralph Barone
wrote:

John B. wrote:
On Tue, 7 Jul 2020 00:52:05 +0000 (UTC), Ralph Barone
wrote:

John B. wrote:
On Mon, 6 Jul 2020 10:41:58 -0700 (PDT), Lou Holtman
wrote:

On Monday, July 6, 2020 at 7:15:32 PM UTC+2, Mark J. wrote:



It's a topic for another post, but I find on the road that energy /per
mile/ is surprisingly consistent, ranging from around 27 kJ/mile for a
gentle pootle to maybe 35 kJ/mile for a very brisk hilly ride, with most
rides in the 30-33 range. Not, of course, on rides that start and end
at different elevations.


Tell that to the people that say that 10-20 Watt on average is insignificant.

Lou

Well, it is the difference between 0.01 and 0.02 H.P. :-)
--
Cheers,

John B.



If 1000 W = 1 HP, yes...

I believe that the usual calculation is 1 hp = 745.69987 watts.
--
Cheers,

John B.



That’s what I thought too, but 1 HP = 1000 W was the only way to explain
your previous post. Maybe you’ve got better horses over there :-)


Well granted I did round a fairly lengthy number but no, 1 hp /
745.69987 watts = 0.0013410220896 hp.
So 10 watts is actual 0.013410220896 hp and 20 watts = 0.0268204417919
hp



Are you SURE about that last significant figure? After all, this is
regarding bicycling, where every minute change is critical! ;-)


Ah well, my calculator were "rounded" a bit so perhaps my numbers are
a bit off. So in the pursuit of greater accuracy one watt should be
referenced as 0.00134102209244049000 hp.


That's better! Thank you.


--
- Frank Krygowski
  #56  
Old July 8th 20, 07:05 AM posted to rec.bicycles.tech
Ralph Barone[_4_]
external usenet poster
 
Posts: 853
Default Fat tire riders look like "fat heads."

Frank Krygowski wrote:
On 7/7/2020 8:46 PM, John B. wrote:
On Tue, 7 Jul 2020 12:19:55 -0400, Frank Krygowski
wrote:

On 7/7/2020 1:23 AM, John B. wrote:
On Tue, 7 Jul 2020 03:56:54 +0000 (UTC), Ralph Barone
wrote:

John B. wrote:
On Tue, 7 Jul 2020 00:52:05 +0000 (UTC), Ralph Barone
wrote:

John B. wrote:
On Mon, 6 Jul 2020 10:41:58 -0700 (PDT), Lou Holtman
wrote:

On Monday, July 6, 2020 at 7:15:32 PM UTC+2, Mark J. wrote:



It's a topic for another post, but I find on the road that energy /per
mile/ is surprisingly consistent, ranging from around 27 kJ/mile for a
gentle pootle to maybe 35 kJ/mile for a very brisk hilly ride, with most
rides in the 30-33 range. Not, of course, on rides that start and end
at different elevations.


Tell that to the people that say that 10-20 Watt on average is insignificant.

Lou

Well, it is the difference between 0.01 and 0.02 H.P. :-)
--
Cheers,

John B.



If 1000 W = 1 HP, yes...

I believe that the usual calculation is 1 hp = 745.69987 watts.
--
Cheers,

John B.



That’s what I thought too, but 1 HP = 1000 W was the only way to explain
your previous post. Maybe you’ve got better horses over there :-)


Well granted I did round a fairly lengthy number but no, 1 hp /
745.69987 watts = 0.0013410220896 hp.
So 10 watts is actual 0.013410220896 hp and 20 watts = 0.0268204417919
hp


Are you SURE about that last significant figure? After all, this is
regarding bicycling, where every minute change is critical! ;-)


Ah well, my calculator were "rounded" a bit so perhaps my numbers are
a bit off. So in the pursuit of greater accuracy one watt should be
referenced as 0.00134102209244049000 hp.


That's better! Thank you.



I feel faster already.

  #57  
Old July 8th 20, 05:41 PM posted to rec.bicycles.tech
Jeff Liebermann
external usenet poster
 
Posts: 4,018
Default Fat tire riders look like "fat heads."

On Tue, 7 Jul 2020 20:04:56 +0000 (UTC), Ralph Barone
wrote:

Frank Krygowski wrote:
On 7/6/2020 10:19 PM, Ralph Barone wrote:
Jeff Liebermann wrote:
On Tue, 07 Jul 2020 07:20:22 +0700, John B.
wrote:

Doesn't wind resistance increase as a square of the velocity?

Yep. Wind resistance is drag and is proportional to the square of the
velocity. However, the energy needed to overcome that drag increases
with the cube of the velocity:
https://physics.info/drag/
See the section on "drag and power":
...if drag is proportional to the square of speed,
then the power needed to overcome that drag is proportional
to the cube of speed (P ? v3). You want to ride your
bicycle twice as fast, you'll have to be eight times
more powerful. This is why motorcycles are so much
faster than bicycles.


Right, but you get there in half the time, so the energy required to go a
certain distance should go up as the square of the speed, once you reach a
velocity where aerodynamic drag dominates.


I'd leave Time out of it.

I'd just say that Energy, like Work, is Force times Distance. If you
were to double your speed, the Force required would indeed increase by a
factor of 4 (i.e. the square of speed, as you said). Multiplying that by
the same distance traveled yields four times the Work (or Energy) required.

I'm sure there are other ways of thinking about this, but my explanation
seems simplest to me.


Yes, thats the clean sheet of paper approach, but I was trying to get
from Jeffs post back to Marks conclusion.


I've spend the last day going round and round trying to make sense of
this. I didn't think about halving the distance when I wrote my
reply. As such, that makes sense. Twice as fast requires 4 times the
energy. Maybe a Gedankenexperiment will help:

I spent some time dealing with wind turbines, where the rule of thumb
is in order to obtain twice the power output requires 8 times the wind
velocity. That's what happens when we take distance traveled out of
the problem.
https://home.uni-leipzig.de/energy/energy-fundamentals/15.htm
The wind power increases with the cube of the wind speed.
In other words: doubling the wind speed gives eight times
the wind power.
I then assumed that the cube rule would also work in the opposite
direction. To move the wind twice as fast requires eight times the
energy driving the motor and turbine. Again, notice that there's no
distance involved, yet. However, if I place a bicycle in the air flow
with a sail to catch the wind, the bicycle would be traveling at the
same speed as the wind. The bicycle will still arrive in half the
time, but because the 8x energy had already been delivered to the wind
by the motor and turbine, it still requires eight time the energy to
product double the velocity and half the a travel time.

This is where I'm stuck. Both Ralph's explanation and hopefully
thought experiment seem to make sense, yet are contradictory. Occam's
Razor favors Ralph's explanation. I'll dig through Bicycle Science by
David Gordon Wilson and see if I make some sense of the problem.

Sorry for the delayed reply. I wanted to think about it and was also
busy getting my Subaru fixed and smogged.


--
Jeff Liebermann
150 Felker St #D
http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558
  #58  
Old July 8th 20, 05:55 PM posted to rec.bicycles.tech
Frank Krygowski[_4_]
external usenet poster
 
Posts: 10,538
Default Fat tire riders look like "fat heads."

On 7/8/2020 12:41 PM, Jeff Liebermann wrote:
On Tue, 7 Jul 2020 20:04:56 +0000 (UTC), Ralph Barone
wrote:

Frank Krygowski wrote:
On 7/6/2020 10:19 PM, Ralph Barone wrote:
Jeff Liebermann wrote:
On Tue, 07 Jul 2020 07:20:22 +0700, John B.
wrote:

Doesn't wind resistance increase as a square of the velocity?

Yep. Wind resistance is drag and is proportional to the square of the
velocity. However, the energy needed to overcome that drag increases
with the cube of the velocity:
https://physics.info/drag/
See the section on "drag and power":
...if drag is proportional to the square of speed,
then the power needed to overcome that drag is proportional
to the cube of speed (P ? v3). You want to ride your
bicycle twice as fast, you'll have to be eight times
more powerful. This is why motorcycles are so much
faster than bicycles.


Right, but you get there in half the time, so the energy required to go a
certain distance should go up as the square of the speed, once you reach a
velocity where aerodynamic drag dominates.


I'd leave Time out of it.

I'd just say that Energy, like Work, is Force times Distance. If you
were to double your speed, the Force required would indeed increase by a
factor of 4 (i.e. the square of speed, as you said). Multiplying that by
the same distance traveled yields four times the Work (or Energy) required.

I'm sure there are other ways of thinking about this, but my explanation
seems simplest to me.


Yes, that’s the “clean sheet of paper” approach, but I was trying to get
from Jeff’s post back to Mark’s conclusion.


I've spend the last day going round and round trying to make sense of
this. I didn't think about halving the distance when I wrote my
reply. As such, that makes sense. Twice as fast requires 4 times the
energy. Maybe a Gedankenexperiment will help:

I spent some time dealing with wind turbines, where the rule of thumb
is in order to obtain twice the power output requires 8 times the wind
velocity. That's what happens when we take distance traveled out of
the problem.
https://home.uni-leipzig.de/energy/energy-fundamentals/15.htm
The wind power increases with the cube of the wind speed.
In other words: doubling the wind speed gives eight times
the wind power.
I then assumed that the cube rule would also work in the opposite
direction. To move the wind twice as fast requires eight times the
energy driving the motor and turbine. Again, notice that there's no
distance involved, yet. However, if I place a bicycle in the air flow
with a sail to catch the wind, the bicycle would be traveling at the
same speed as the wind. The bicycle will still arrive in half the
time, but because the 8x energy had already been delivered to the wind
by the motor and turbine, it still requires eight time the energy to
product double the velocity and half the a travel time.

This is where I'm stuck. Both Ralph's explanation and hopefully
thought experiment seem to make sense, yet are contradictory. Occam's
Razor favors Ralph's explanation. I'll dig through Bicycle Science by
David Gordon Wilson and see if I make some sense of the problem.


I'll say this on first thought, without taking time now to think about
it in more detail.

Your problem may be that you're conflating energy and power. Check to
see if your bike needs 8 times as much power, but only four times as
much energy (i.e. 4 times as much work).

Gotta go.


--
- Frank Krygowski
  #59  
Old July 8th 20, 10:39 PM posted to rec.bicycles.tech
Jeff Liebermann
external usenet poster
 
Posts: 4,018
Default Fat tire riders look like "fat heads."

On Wed, 08 Jul 2020 07:46:16 +0700, John B.
wrote:

Ah well, my calculator were "rounded" a bit so perhaps my numbers are
a bit off. So in the pursuit of greater accuracy one watt should be
referenced as 0.00134102209244049000 hp.


Umm... which type of horsepower is that?
https://en.wikipedia.org/wiki/Horsepower
Imperial, mechanical, metric, electrical, boiler, hydraulic, or air
HP?

--
Jeff Liebermann
150 Felker St #D
http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558
  #60  
Old July 9th 20, 12:36 AM posted to rec.bicycles.tech
Radey Shouman
external usenet poster
 
Posts: 1,747
Default Fat tire riders look like "fat heads."

Jeff Liebermann writes:

On Tue, 7 Jul 2020 20:04:56 +0000 (UTC), Ralph Barone
wrote:

Frank Krygowski wrote:
On 7/6/2020 10:19 PM, Ralph Barone wrote:
Jeff Liebermann wrote:
On Tue, 07 Jul 2020 07:20:22 +0700, John B.
wrote:

Doesn't wind resistance increase as a square of the velocity?

Yep. Wind resistance is drag and is proportional to the square of the
velocity. However, the energy needed to overcome that drag increases
with the cube of the velocity:
https://physics.info/drag/
See the section on "drag and power":
...if drag is proportional to the square of speed,
then the power needed to overcome that drag is proportional
to the cube of speed (P ? v3). You want to ride your
bicycle twice as fast, you'll have to be eight times
more powerful. This is why motorcycles are so much
faster than bicycles.


Right, but you get there in half the time, so the energy required to go a
certain distance should go up as the square of the speed, once you reach a
velocity where aerodynamic drag dominates.


I'd leave Time out of it.

I'd just say that Energy, like Work, is Force times Distance. If you
were to double your speed, the Force required would indeed increase by a
factor of 4 (i.e. the square of speed, as you said). Multiplying that by
the same distance traveled yields four times the Work (or Energy) required.

I'm sure there are other ways of thinking about this, but my explanation
seems simplest to me.


Yes, that’s the “clean sheet of paper” approach, but I was trying to get
from Jeff’s post back to Mark’s conclusion.


I've spend the last day going round and round trying to make sense of
this. I didn't think about halving the distance when I wrote my
reply. As such, that makes sense. Twice as fast requires 4 times the
energy. Maybe a Gedankenexperiment will help:

I spent some time dealing with wind turbines, where the rule of thumb
is in order to obtain twice the power output requires 8 times the wind
velocity. That's what happens when we take distance traveled out of
the problem.
https://home.uni-leipzig.de/energy/energy-fundamentals/15.htm
The wind power increases with the cube of the wind speed.
In other words: doubling the wind speed gives eight times
the wind power.
I then assumed that the cube rule would also work in the opposite
direction. To move the wind twice as fast requires eight times the
energy driving the motor and turbine. Again, notice that there's no
distance involved, yet. However, if I place a bicycle in the air flow
with a sail to catch the wind, the bicycle would be traveling at the
same speed as the wind. The bicycle will still arrive in half the
time, but because the 8x energy had already been delivered to the wind
by the motor and turbine, it still requires eight time the energy to

^^^^^^
There's your problem -- it requires eight times the *power*, but only
four times the total energy. Power is energy (or work) divided by time. If
you quadruple the force retarding your bicycle, and also double the
speed (distance divided by time), then you'll have to deliver eight
times the power (distance multiplied by force divided by time).

product double the velocity and half the a travel time.



This is where I'm stuck. Both Ralph's explanation and hopefully
thought experiment seem to make sense, yet are contradictory. Occam's
Razor favors Ralph's explanation. I'll dig through Bicycle Science by
David Gordon Wilson and see if I make some sense of the problem.

Sorry for the delayed reply. I wanted to think about it and was also
busy getting my Subaru fixed and smogged.


--
 




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