Talk:Bicycle performance

message: It might be interesting to mention "mile-a-minute Murphy" here, who went 62 mph (100 kph) in 1899:  I'm not sure how notable this is though. --NE2 08:24, 22 January 2007 (UTC)
replies: text: Not very notable, at least in the context of this article.  Motor-pacing virtually eliminates aerodynamic drag. Mr Larrington (talk) 12:18, 23 July 2009 (UTC)
thread_title: Mile-a-minute Murphy

message: The "These are actually kilocalories" statement seems unnecessarily confusing. It would be better to use either "calories" or "kilocalories" correctly or not at all. -AndrewDressel 20:29, 22 January 2007 (UTC)
replies: text: Done, finally, 11 months later. -AndrewDressel (talk) 20:44, 28 December 2007 (UTC)
thread_title: Calories vs kilocalories

message: Fram:  As you said, this isn't your area of expertise.  The type of calculations performed in this article are completely routine back-of-the-envelope analysis that physics students are taught to perform in their first year of college.  They're not original research.  As for whether this subject could be referenced.  Ahem.    etc.
This material does need to be shortened, revised for tone and moved though. -- pde 09:33, 3 February 2007 (UTC)
replies: text: As I said, I don't mean referencing the formulas (which are indeed routine ones), but the article, the subject matter as an independent field of research. I think you have misread my statement there. But thanks for the references for the subject, they can probably be added to the article (if they are WP:RS soutces). Fram 20:28, 3 February 2007 (UTC)
thread_title: References needed?

message: The claim that the KE for the rotational component is equal to the moving componentis not correct.  The claim is only true if ALL of the wheels mass is at the perimeter.  The rotational velocity of the wheel at the hub is much smaller, and the hub/spokes are a significant portion of the total mass of the wheel. The rotational velocity is a function of the radius -- v(r).  To correctly calculate the rotational KE, you would have to perform an integration over the entire radius of the wheel.  In any case the rotational component will be a fraction of the moving component...so the notion that "A pound off the wheels = 2 pounds off the frame." is (at least mathematically) false unless you took that pound entirely from the tires. 13:08, 27 April 2007 (UTC)
replies: text: Yes, it is an approximation, but not a bad one. A quick search online finds a Dura-Ace front hub with 129g, a DT Swiss rim with 415g, a Continental GP4000 tire with 205g, and a tube with 55g. 32 spokes are abaout 300g. So, of the total mass, over 60% is close to the bead diameter. The spokes are less than 30% and the hub less than 12%. Low-spoke-count wheels make the approximation even better. -AndrewDressel 19:18, 27 April 2007 (UTC)
thread_title: Rotating kinetic energy, need to do some calculus

message: The translational kinetic energy of an object in motion is:
:E = \tfrac{1}{2}mv^2,
Where E is energy in joules, m is mass in kg, and v is velocity in meters per second.  For a rotating mass (such as a wheel), the rotational kinetic energy is given by
:E = \tfrac{1}{2}I \omega^2,
where I is the moment of inertia, \omega is the angular velocity in radians per second, r is the radius in meters.  For a wheel with all its mass at the radius (a fair approximation for a bicycle wheel), the moment of inertia is
:I = m r^2.
The angular velocity is related to the translational velocity and the radius of the tire.  As long as there is no slipping,
:\omega = \frac{v}{r}.
When a rotating mass is moving down the road, its total kinetic energy is the sum of its translational kinetic energy and its rotational kinetic energy:
:E = \tfrac{1}{2}mv^2 + \tfrac{1}{2}I\omega^2
Substituting for I and \omega, we get
:E = \tfrac{1}{2}mv^2 + \tfrac{1}{2}mr^2 \cdot \frac{v^2}{r^2} 
The r^2 terms cancel, and we finally get
:E = \tfrac{1}{2}mv^2 + \tfrac{1}{2}mv^2 = mv^2.
In other words, a mass on the tire has twice the kinetic energy of a
non-rotating mass on the bike.  There is a kernel of truth in the old saying
that "A pound off the wheels = 2 pounds off the frame."
:This math is in error! the r in the moment of enertia formula is for the radius of an approximating point mass, not the outside of the wheel. thus the velocity of that mass is not the same as the velocity of the ground speed. They do not cancel! -pvd (130.212.215.215)
::The formula is correct for a ring.  For a ball it would be 2/5 times that, so I think the maths is correct and, more importantly, so is the physics -ejf (194.30.186.250)
:::I agree with ejf. The formula is correct. Its usefulness however depends on how well a thin hoop approximates a particular bicycle wheel. This, I believe is pvd's objection. In reality, all the mass cannot be at the radius. For comparison, the opposite extreme might be a disk wheel where the mass is distributed evenly throughout the interior. In this case I = \tfrac{1}{2}m r^2 and so the resulting total kinetic energy becomes E = \tfrac{1}{2}mv^2 + \tfrac{1}{4}mv^2 = \tfrac{3}{4}mv^2. A pound off the disk wheels = only 1.5 points off the frame. Most real bicycle wheels will be somewhere between these two extremes. -AndrewDressel (talk) 22:51, 18 January 2008 (UTC)
thread_title:  Kinetic Energy 

message: This article doesn't seem to have any mention of drag/rolling resistance due to the tyres; underinflated tyres are widely beleived to cause a lot of drag, but are there any useful references for this effect? Murray Langton (talk) 12:25, 18 July 2011 (UTC)
thread_title:  effect of bicycle tyres 

message: In relation to the following extract:
Power required
There is a well known equation that gives the power required to push a bike/rider through the air and to overcome the friction of the drive train:
:P = g m V_g (K_1+s) + K_2 \times V_a^2 V_g
Where P is in watts, g is Earth's gravity, Vg is ground speed (m/s), m is bike/rider mass in kg, s is the grade (m/m), and Va is the rider's speed through the air (m/s). K1 is a lumped constant for all frictional losses (tires, bearings, chain), and is generally reported with a value of 0.0053. K2 is a lumped constant for aerodynamic drag and is generally reported with a value of 0.185 kg/m.[16]
Note that the power required to overcome friction and gravity is proportional only to rider weight and ground speed...
I know that this formula appears to come from a credible source (e.g. http://www.sportsci.org/jour/9804/dps.html#ref ), but it doesn't make sense to me that the power required to overcome gravity is proportional to speed. (Perhaps the original author has confused 'work done' with 'force required'?)
This formula suggests that the power required to overcome gravity at zero speed is zero, when in fact isn't gravity a constant downward force accelerating an object by 9.81m/s? On a flat, natural force counteracts gravity, but on a slope, this net force is a function of massive, gravity, and slope angle (specifically m.g.sin(angle) - see http://www.studyphysics.ca/newnotes/20/unit01_kinematicsdynamics/chp06_vectors/lesson25.htm )
I suggest the formula be amended accordingly. Anyone care to comment on this?
:Since mechanical power is defined to be the dot product of the force vector and the velocity vector, the power is zero if the veloity is zero. On a slope, the force required to remain stationary may be non-zero, but the mechanical power required is zero. -AndrewDressel (talk) 16:17, 20 August 2011 (UTC)
thread_title:  Power required to overcome Gravity 

message: The section on kinetic energy refers to the mass of the wheel, and seems to be using the weight in the equations. Could someone take a look and comment. If this error has been made it will not change the conclusion that a kg or wheel is equivalent to 1.5-2 kg of frame, but the numbers themselves will be much smaller.  — Preceding unsigned comment added by M610 (talk • contribs) 17:45, 19 November 2011 (UTC)
thread_title:  Weight, or Mass? 

message: I seem to be profoundly confused - perhaps someone can help me out here.  Say I weigh 100 kg and go for a 1-hr ride in which I travel 30 km, hence 30 km/h.  There are no hills and no wind.  I'd like to know how much energy I need.
Section 1: 1.62 kJ/(km∙kg) = 4860 kJ = 1161 kcal, compatible with what I find on many websites (on the order of 1000 kcal for ~1h).
Section 4.1: using the formula, I get P = 43 + 107 = 150 W, which is compatible with the 175 W example below the formula and commentators on the Tour de France claiming thesr guys do ~250 W.  Biking for 1h, I need 150 W ∙ 3600 s = 540 kJ = 129 kcal.
How can the difference of about an order of magnitude(!) between the rules of thumb "a few 100 W in power" on the one hand and "on the order of 1000 kcal in 1h" on the other be explained?
AstroFloyd (talk) 22:20, 8 July 2012 (UTC)
replies: text: OK, I guess there's a factor of four in the efficiency of the human body, as stated further down in Sect. 4.1: "The human body runs at about 24% efficiency for a relatively fit athlete, so for every kilojoule delivered to the pedals the body consumes 1 kcal (4.2 kJ) of food energy."  I suppose that brings the difference down to a factor of two... AstroFloyd (talk) 07:46, 9 July 2012 (UTC)
thread_title:  Energy/power needed for cycling??? 

message: The article should start with the "Power Required" section which is relevant and very useful.
The whole thing about the KE of the wheels be put later or in a footnote or another page,
because is actual relevance to the efficiency of a bike is very small.
That is, all that massive discourse is just about the energy which you put in to
get the bike going once when you start, which for most rides and most races is a very very small part
of the total energy, and which you in fact can get back if you rest as you let the bike come to a halt at the end of the ride.
I have found people (who may have read this) who believe it is important to get the weight of the rims down,
compared to the fixed parts of the bike, as though it helped you go faster up hills or steadily along the flat. 
So the whole KE section could do with some sort of  disclaimer on it.
The KE of the wheels is irrelevant to the performance a bike being ridden steadily once it is going.  TimBL (talk)  —Preceding undated comment added 18:59, 9 July 2012 (UTC)
replies: text: I've tried to quantify this and removed the excessively long derivation. Theosch (talk) 10:34, 30 March 2015 (UTC)
thread_title:  Wheel k.e. mumbo jumbo 

message: With sentences like "possible technical explanations"...  "placebo effect" and so forth which shows it was guesses and invented contents by editors. Cantaloupe2 (talk) 10:12, 7 December 2012 (UTC)
thread_title:  'explanations' trimmed off 

message: anyone up on frame flex and wheel hub lateral and torsional flex ? I would love to see a reference link to a slow motion shoot  of a hub shifting around on the spokes - this must be out there somewhere--— ⦿⨦⨀Tumadoireacht Talk/Stalk 09:29, 31 December 2013 (UTC)
thread_title:  flex 

message: Should this line
"In other words, a mass on the tire has twice the kinetic energy of a non-rotating mass on the bike. This all depends, of course, on how well a thin hoop approximates the bicycle wheel. In reality, all the mass cannot be at the radius." 
correctly read 
In other words, a mass on the tire has twice the kinetic energy of a non-rotating mass on the bike. This all depends, of course, on how well a thin hoop approximates the bicycle wheel. In reality, all the mass cannot be at the rim/outer edge
--— ⦿⨦⨀Tumadoireacht Talk/Stalk 09:15, 18 March 2015 (UTC)
replies: text: I think this is what is meant, but I agree that it is confusing. Theosch (talk) 16:30, 29 March 2015 (UTC)
thread_title:  Reduction of weight vs reduction of rotating mass/Kinetic energy of a rotating wheel 

message: The section "Climbing Power" contains this claim: "As this power is used to increase the potential energy of bike and rider, it is returned when going downhill and not lost unless the rider brakes or travels faster than desired."  
Ahm, I don't think so.  The potential energy is converted to kinetic energy of the bicycle and rider, plus kinetic energy of the air displaced as the rider speeds down the hill (and eventually into heat as the vortices dissipate), plus friction in the bearings and tires and other things that move.  If the downhill leg is a very shallow grade, and the downhill leg consequently slow, more kinetic energy will go into the bike-and-rider, and less into turbulence (because turbulence is non-linear).  Only as his downhill speed goes to zero will all the potential energy at the top be converted to the bike-and-rider's kinetic energy.  Nature just isn't fair, is she?
This means that a hilly course takes more of the rider's energy, for a given speed, than a flat course: the energy put into climbing is never fully recovered.  I ride with someone who is entirely convinced the speeds he attains on the downhill portions more than make up for the work required to climb.  I don't believe it, and am pretty sure he has no corroborative data.  (I should ask him to coast up the next equally high hill.)
But I have made no edit, in case this turns out to be contested. Captain Puget (talk) 06:00, 24 November 2015 (UTC)
replies: text: The potential energy is fully recovered when the original altitude is reached but you are correct that it is converted to heat except for any surplus bit of kinetic energy at the bottom. I've reworded slightly in order to hopefully clarify. Theosch (talk) 12:42, 12 July 2016 (UTC)
thread_title:  Potential energy regained? 

message: The last sentence of the lead seems to be ambiguous: "In terms of the ratio of cargo weight a bicycle can carry to total weight, it is also a most efficient means of cargo transportation." If it's only "a" most efficient means, the sentence would read better as "...it is also an efficient means...", but if we're claiming it's "the" most efficient means, the sentence should be written that way (and be cited).  — Preceding unsigned comment added by Kajabla (talk • contribs) 20:41, 11 July 2016 (UTC)
thread_title:  Lead section wording 

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