A better argument for today - how much weight matters
#79
Senior Member
Popularly known as "watts."
Ignoring everything but gravity and keeping power constant, time is linear with mass. So your method is fine. Cutting weight in half cuts climbing time in half. Cutting weight to zero places you at both the top and bottom of the hill at the same time.
Ignoring everything but gravity and keeping power constant, time is linear with mass. So your method is fine. Cutting weight in half cuts climbing time in half. Cutting weight to zero places you at both the top and bottom of the hill at the same time.
#80
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Popularly known as "watts."
Ignoring everything but gravity and keeping power constant, time is linear with mass. So your method is fine. Cutting weight in half cuts climbing time in half. Cutting weight to zero places you at both the top and bottom of the hill at the same time.
Ignoring everything but gravity and keeping power constant, time is linear with mass. So your method is fine. Cutting weight in half cuts climbing time in half. Cutting weight to zero places you at both the top and bottom of the hill at the same time.
#84
out walking the earth
Thread Starter
crashed on the bell lap, sitting in the four hole and certain I'd win the field sprint. going to become a hill climb specialist, so carry on.
#85
Making a kilometer blurry
That really blows. You and shovel doing alright? Sucks when guys like you go down, because I know you weren't just overreacting to something in front of you like 99% of the crash videos posted around here. Well, and it sucks for anyone to crash.
#86
out walking the earth
Thread Starter
In tge the interest of fairness some times things just happen I crashed on tge bell lap at the pointy end of the field (2 away). A bunch of us were going aggressive for the same small space. The rider I collided with cdt's teammate was up against tge barrier and had no where to go. He saw it as my fault. Of course I didn't think he belonged where he was.
#88
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The following is intended respectfully.
It’s unnecessary to compute the additional element of “force”, as there’s a more direct method being that you already know the three key factors, those being, the object’s mass, the gain in vertical elevation (height), and gravity’s accelerative influence, 9.8 m/s/s.
In such cases, we more simply apply the straightforward longstanding equation associated with “gravitational potential energy” as provided below:
E = mgh
Whereby,
E = gravitational potential energy “in Joules”
m = mass of object “in kg”
g = acceleration via gravity, a constant 9.8 m/s/s
h = height to which object must be raised “in meters”
.25 pounds * .45359237 = .113398093 kg mass
3,000 feet * .3048 = 914.4 meters height
Once the conversion to SI (Standard International) units has been made, we apply these factors to the gravitational potential energy equation, E = mgh, to yield the product, Joules of energy:
.113398093 kg * 9.8 m/s/s * 914.4 meters = 1016.173915 Joules of energy
If we wish to derive the average power required per second during the ascent, we merely divide the Joules of energy by the total seconds required to achieve the desired gain in vertical elevation (height). For instance, if it requires 1 hour, we divide by 3,600 seconds:
1016.173915 Joules / 3,600 seconds = 0.282270532 Joules of energy (per second)
Thus, the element of “force” need not be computed. Naturally, it’s anyone’s prerogative to compute the desired outcome it in any manner that yields the correct answer however, this is the most direct method in physics concerning this matter.
It’s unnecessary to compute the additional element of “force”, as there’s a more direct method being that you already know the three key factors, those being, the object’s mass, the gain in vertical elevation (height), and gravity’s accelerative influence, 9.8 m/s/s.
In such cases, we more simply apply the straightforward longstanding equation associated with “gravitational potential energy” as provided below:
E = mgh
Whereby,
E = gravitational potential energy “in Joules”
m = mass of object “in kg”
g = acceleration via gravity, a constant 9.8 m/s/s
h = height to which object must be raised “in meters”
.25 pounds * .45359237 = .113398093 kg mass
3,000 feet * .3048 = 914.4 meters height
Once the conversion to SI (Standard International) units has been made, we apply these factors to the gravitational potential energy equation, E = mgh, to yield the product, Joules of energy:
.113398093 kg * 9.8 m/s/s * 914.4 meters = 1016.173915 Joules of energy
If we wish to derive the average power required per second during the ascent, we merely divide the Joules of energy by the total seconds required to achieve the desired gain in vertical elevation (height). For instance, if it requires 1 hour, we divide by 3,600 seconds:
1016.173915 Joules / 3,600 seconds = 0.282270532 Joules of energy (per second)
Thus, the element of “force” need not be computed. Naturally, it’s anyone’s prerogative to compute the desired outcome it in any manner that yields the correct answer however, this is the most direct method in physics concerning this matter.
#89
out walking the earth
Thread Starter
so 1.25-1.5 pounds in the wheels...is that more significant than similar weight somewhere else, and and in real world terms of time and dollars and sense what's it worth?
#90
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It's very very slightly worth more than weight elsewhere. Not sure if it's worth the value without some numbers.
#92
Making a kilometer blurry
I ran numbers based on actual 4-corner crit accelerations. If you completely removed the weight of decent tires and tubes from the wheels, it was something like 1/5 of a Watt average for the race, just from rotational accelerations -- which is the only difference from carrying the weight elsewhere. Since some of this rotating weight might be in the hub too, where it has much less torsional influence, 1/5 of a Watt is an unlikely-to-be-met maximum.
If you're on a course with little or no braking, then it's probably a wash, as the rotational inertia will help you carry speed into a hill.
If you're on a course with little or no braking, then it's probably a wash, as the rotational inertia will help you carry speed into a hill.
#93
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yep, what WR says is correct. there's a difference but it is so small as to not be worth considering.
in most cases, aero trumps weight. most = when grades are below 6% and when one rides down whatever was ridden up. for special cases (finishing climbs are pure HCs with grades 6% or steeper) weight starts to get a small advantage.
the amount varies with rider weight and power, but at "normal" racer weights and power the above is true. the lighter/faster you are, the greater the benefit of aero gear. i.e., chris froome can benefit from aero equipment up steeper grades than i can.
in most cases, aero trumps weight. most = when grades are below 6% and when one rides down whatever was ridden up. for special cases (finishing climbs are pure HCs with grades 6% or steeper) weight starts to get a small advantage.
the amount varies with rider weight and power, but at "normal" racer weights and power the above is true. the lighter/faster you are, the greater the benefit of aero gear. i.e., chris froome can benefit from aero equipment up steeper grades than i can.
#94
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yep, what WR says is correct. there's a difference but it is so small as to not be worth considering.
in most cases, aero trumps weight. most = when grades are below 6% and when one rides down whatever was ridden up. for special cases (finishing climbs are pure HCs with grades 6% or steeper) weight starts to get a small advantage.
the amount varies with rider weight and power, but at "normal" racer weights and power the above is true. the lighter/faster you are, the greater the benefit of aero gear. i.e., chris froome can benefit from aero equipment up steeper grades than i can.
in most cases, aero trumps weight. most = when grades are below 6% and when one rides down whatever was ridden up. for special cases (finishing climbs are pure HCs with grades 6% or steeper) weight starts to get a small advantage.
the amount varies with rider weight and power, but at "normal" racer weights and power the above is true. the lighter/faster you are, the greater the benefit of aero gear. i.e., chris froome can benefit from aero equipment up steeper grades than i can.
#95
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So...in general, for the average cyclist/racer, it's better (in terms of performace, $$, and watts saved) to lose 5 pounds from the body than 0.5-1 pound from the bike? I'm starting to feel like this bike industry has duped me.
#96
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in most cases, aero trumps weight. most = when grades are below 6% and when one rides down whatever was ridden up. for special cases (finishing climbs are pure HCs with grades 6% or steeper) weight starts to get a small advantage.
the amount varies with rider weight and power, but at "normal" racer weights and power the above is true. the lighter/faster you are, the greater the benefit of aero gear. i.e., chris froome can benefit from aero equipment up steeper grades than i can.
the amount varies with rider weight and power, but at "normal" racer weights and power the above is true. the lighter/faster you are, the greater the benefit of aero gear. i.e., chris froome can benefit from aero equipment up steeper grades than i can.
I still went with reynolds rims over zipp for the 200g weight advantage because I'm a sucker (and 1/3 the cost and relatively minor aero delta to boot)
#97
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I really agree with the above logic. I just wish I saw what I believe in practice more often. Granted, it is good if competition doesn't catch on but...
1) why don't I see more skinsuits in crits?
2) shoe covers?
3) deeper wheels? Our main local crit is not technical at all. One wide 180 is about the only real braking needed and to that degree I would rock stinger 90/70 combo or similar if I had the funds.
What gets interesting @Ygduf is not only the weight savings of a wheel like reynolds but how each company decides on what yaw angle to focus on. I think over the course of a crit you won't see much past 10 degrees in a typical paceline and often wonder why more companies don't focus on low yaw angles. HED for example is VERY good at low yaw angles
#98
out walking the earth
Thread Starter
Let's tip the hill above 6%. Make it 8, for an hour. What's a pound save, and what $ amount would you be willing to spend under what circumstances?
#99
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I really agree with the above logic. I just wish I saw what I believe in practice more often. Granted, it is good if competition doesn't catch on but...
1) why don't I see more skinsuits in crits?
2) shoe covers?
3) deeper wheels? Our main local crit is not technical at all. One wide 180 is about the only real braking needed and to that degree I would rock stinger 90/70 combo or similar if I had the funds.
1) why don't I see more skinsuits in crits?
2) shoe covers?
3) deeper wheels? Our main local crit is not technical at all. One wide 180 is about the only real braking needed and to that degree I would rock stinger 90/70 combo or similar if I had the funds.
#100
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I don't have a specific enough scenario in mind to opine about the dollar value of saving that time.
Most of my opinion on this matter is informed by a single race with a 10 minute, 4% varied hill, where I chose heavier aero wheels, fell off the breakaway by just a few seconds on the climb, and couldn't get back on. A pound off the wheels might have kept me in the break.
Last edited by globecanvas; 08-08-14 at 02:13 PM.