## Is my maths correct?? Climbing and w/kg

Discuss light weight issues concerning road bikes & parts.
forardun
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Joined: Sun Jan 15, 2017 11:59 am
shinzomaeda wrote:

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so..... how much faster will ceramic derailleur jockey bearings make me?

haynesdavid1988
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Joined: Fri Jan 20, 2017 8:39 am
Marin wrote:
RimClencher wrote:Something none of these calculator sites seems to mention is the assumption you are riding at constant velocity. Almost no one is able to maintain an absolute constant velocity riding uphill. Even the pros tend to yo-yo along depending on where they are in the pedal stroke. Each small deceleration requires a small acceleration, which takes more energy than riding at constant velocity. These decelerations get more pronounced the steeper the gradient, so the steeper the gradient the less accurate these online calculators, and your calculations, will turn out to be.

If you follow your own train of thought, the decelerations and accelerations will cancel out. Where should the energy go?

Exactly. When climbing your power is generally within 5 percent of you pace it properly which is why you can calculate it on best bike split etc. And it does work and it is accurate. And is a good tool for things like this and kit selection.

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Marin
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forardun wrote:so..... how much faster will ceramic derailleur jockey bearings make me?

That's easy - exactly 0%

Marin
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shinzomaeda wrote:Yes, sorry, I was being very lazy. Is chart relevant or helpful?

It's very helpful - I was lazy, you did the work for me - thanks!

basurper
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How about the crank arm length? Will the 170mm need less watt than 165mm?

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Marin
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No, of course not. Do you know what Watt is?

jpanspac
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Joined: Thu Apr 11, 2013 8:21 pm
.
My favorite components are the ones I never have to think about.

jpanspac
Posts: 160
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darnellrm wrote:Not at all... The heavier bike will decelerate quicker and accelerate slower each cycle.

No, the heavier bike will be slower to decelerate, and slower to accelerate.

F=M*a, where F is gravity (during the deceleration), and M is the combined mass of the rider and bike.

So a=F/M, which means acceleration (or deceleration) decreases as mass increases.
My favorite components are the ones I never have to think about.

RimClencher
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Marin wrote:If you follow your own train of thought, the decelerations and accelerations will cancel out. Where should the energy go?

Yep, you're right.

sgergole
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Joined: Fri Mar 18, 2016 11:39 am
From my experience and tested on the field, 1kg of weight equals to about 4w. (i weight 67kg and ftp at around 300w). A furhter approximation related to the impact this has on time is that the percentage change in the watts is equal to the percentage change in time needed to complete the climb.

I found that for small changes (up to 4-5 kg) this tends to work quite consistently, at least for me.

Therefore, if I make a 10'=600 seconds climb at 300w with a 7kg bike, and the same climb at 300w with a 6kg bike, I should be able to make a time of 9'52".
On multiple tries, this works for me on various climbs during trainings where i tried to keep the output as stable as possible...

darnellrm
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Location: NC, USA
jpanspac wrote:
darnellrm wrote:Not at all... The heavier bike will decelerate quicker and accelerate slower each cycle.

No, the heavier bike will be slower to decelerate, and slower to accelerate.

F=M*a, where F is gravity (during the deceleration), and M is the combined mass of the rider and bike.

So a=F/M, which means acceleration (or deceleration) decreases as mass increases.

You are neglecting to consider the much greater affect of gravity at the steep inclines that are being discussed.

basilic
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Location: Geneva, Switzerland
respectfully, I think a few points are wrong in this thread.
gravity g is the acceleration that the earth imparts upon an object on its surface. If I recall high school physics, it is the same for all objects, regardless of mass (of course in real life friction due to air will mess it up). g mostly depends on the earth's mass and radius (and the gravitational constant G).
The force varies with the mass of the object, F=mg. So a heavy rider will be subjected to a greater force, and will have to produce a greater force just to keep going (uphill), and since work is force*displacement, it'll cost him more energy (and more power if he wants to keep up with the light rider) to reach the top.
Plus it all depends on the grade of the road, since g acts vertically, and only a component of it holds the rider back, g*sin(artan(grade).

(and apologies if I have only added mistakes; we live in an era of alternate realities after all)

forardun
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Joined: Sun Jan 15, 2017 11:59 am
erm, you wont find anyone arguing that it requires more energy to lift a heavier rider up a hill.

jpanspac
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darnellrm wrote:
jpanspac wrote:
darnellrm wrote:Not at all... The heavier bike will decelerate quicker and accelerate slower each cycle.

No, the heavier bike will be slower to decelerate, and slower to accelerate.

F=M*a, where F is gravity (during the deceleration), and M is the combined mass of the rider and bike.

So a=F/M, which means acceleration (or deceleration) decreases as mass increases.

You are neglecting to consider the much greater affect of gravity at the steep inclines that are being discussed.

The grade of the incline is irrelevant.
My favorite components are the ones I never have to think about.

bm0p700f
in the industry
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Force down the slope =mg*sin (theta) there theta is the angle the slope makes to the horizantal.

arctan(theta) =1/tan(theta)

So I would say gradient is quite important to the force the cyclist feel down the slope and how quickly the accelerate or slow down as power output changes.

Maths is wonderful. Instead of a thread of this and that the only reply needed is a set of equations. They speak for themselves.

Power required to overcome all drag = 0.5*CdA*rho*Va*Vg^2 + Fw*Vg^3+ Crr*mg*Vg + Vg*mg*sin(arctan(%grad/100)

Cd = drag coeff
A frontal area
rho = air density
Va = air speed
Vg = ground speed
Crr = rolling resistance
Fw*Vg = power to rotate wheels aero wheels have a lower value as this is a aerodynamic term for a rotating wheel. Good numbers for Fw are not available aero weenies has a suggested value.

Not related but related at the same time.
PV=nRT

P = air pressure
V = volume
n = no of moles = mass/molar mass = m/Mr
R= gas constant
T = temp in Kelvin

So P*Mr = (m/V)*RT =rho*RT

So rho =P*Mr/RT = P*C/T
C= Mr/R

Power required to overcome all drag = 0.5*CdA*(PC/T)*Va*Vg^2 + +Fw*Vg^3+ Crr*mg*Vg + Vg*mg*sin(arctan(%grad/100)

so as temp drop power needed rises. if you are in a high pressure area 1020mbar vs low pressure 990 mbar you require about 3% more power to over come air resistance.
If I ride in 1000mbar pressure at 20 celcius heat I will find I need another 3.5% more power to overcome air resistance if the temperature was 10 degrees lower.

Lots of factors affect your pace.

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