A better argument for today - how much weight matters
06-19-14, 02:47 PM
#53
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It's feels so liberating not to be a weight weenie. Speaking of which a guy on Weight Weenies did a climb with and without a 25lbs backpack at the same power output. The added weight only cost him seconds over ~20 minutes (if I'm remembering the numbers correctly.)
06-19-14, 03:13 PM
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When it comes down to the cost vs weight I try to convince myself that it'd be easier to just lose the weight off the body instead of from the wallet/bike. That's when I pass on the expensive part and forget to lose the weight off the body.
06-19-14, 03:19 PM
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I'm pretty sure that's bs.
06-19-14, 04:47 PM
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Here's the post, relevant text quoted below. I wasn't exactly remembering the numbers right. Works out to 60 seconds over 20 minutes or around 1 watt of extra power to compensate for every additional pound. Remember that the next time you're tempted to spend a lot of money to shave grams off your bike. Just work on raising your FTP a watt or two or losing weight instead.
It was avg watts for that 5 min segment. So I did like 200 avg watts up the segment. For 5 min. Then 227 watts avg with the (25lb.) backpack on. And the same exact time and speed. I think I went a third time with the backpack at 200 watts and was like 20 seconds or so slower. But that's 25 lbs that's a lot 1-3 lbs is nothing sure...
It was avg watts for that 5 min segment. So I did like 200 avg watts up the segment. For 5 min. Then 227 watts avg with the (25lb.) backpack on. And the same exact time and speed. I think I went a third time with the backpack at 200 watts and was like 20 seconds or so slower. But that's 25 lbs that's a lot 1-3 lbs is nothing sure...
Last edited by Dunbar; 06-19-14 at 04:51 PM.
06-19-14, 08:42 PM
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if someone's well-trained and 8 or 9% body fat, it's not like it's easy to make changes.
even if someone has weight to lose, dropping weight on the bike still makes a difference. whether that difference is of any significance is another matter. (most of the time i think people fixate on weights of bike parts, but i think part of that is that it is easy to understand/quantify differences in grams. aero differences are harder for most to understand.)
06-19-14, 09:32 PM
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Well, the base layer is going to impede evaporation and runoff. If it was going straight to the jersey either it would be evaporating sooner, or the jersey would saturate faster than base layer + jersey, and sweat would start dripping sooner. So, while you did have it in you, once it's not, the base layer keeps it with you longer.
i agree with you re: a traditional base layer.
i have no idea to what degree those new-fangled base layers work in various conditions, but in my limited experience when i've tried one neither my kit nor the base layer wears more than when i started (and those base layers weigh just a handful of grams). perhaps any impeded wicking, as you describe, is countered by improved airflow. dunno.
06-19-14, 09:39 PM
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i'm not the type to ride with my jersey completely unzipped, but for those who do the jersey attached to the shorts means that stuff in one's pockets remains stable even with the jersey unzipped.
06-24-14, 11:05 AM
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What's the total elevation gain on the hill in question?
You're looking at 0.1134kg.
Force required to lift it F=ma, so F = 0.1134kg * 9.8m/s^2 = 1.111N
Work = F * d = 1.111N * elevation gain in meters. For a 3000ft climb, it's 914m, so 1.111N * 914m = 1016Nm = 1016J of work
If you spend an hour climbing this hill, you've 1016J / 60*60s = 0.28J/s = 0.28W to get this 1/4 lb. up the hill at that pace.
So just substitute your elevation in meters and the Strava KOM time for duration and recalculate.
You're looking at 0.1134kg.
Force required to lift it F=ma, so F = 0.1134kg * 9.8m/s^2 = 1.111N
Work = F * d = 1.111N * elevation gain in meters. For a 3000ft climb, it's 914m, so 1.111N * 914m = 1016Nm = 1016J of work
If you spend an hour climbing this hill, you've 1016J / 60*60s = 0.28J/s = 0.28W to get this 1/4 lb. up the hill at that pace.
So just substitute your elevation in meters and the Strava KOM time for duration and recalculate.
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.
06-24-14, 12:36 PM
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someone do the math and tell me how far off this is from my method of taking say 20min on a climb at 196lbs (combined bike/body) weight to be 6.122 seconds/lb, then rounding down and saying if I'm 2 lbs lighter I'll be 12sec faster.
06-24-14, 12:39 PM
#74
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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.
06-24-14, 02:04 PM
#75
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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.