# Two quick effort/workout questions re: efficiency and recovery

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**Two quick effort/workout questions re: efficiency and recovery**

1) Let's say you're on a short climb which, for me, is about 2 minutes long. I give a hard enough, out-of-saddle effort so that my legs are screaming as I go over the top. My instinct is to coast down w/ the least amount of work as possible; however, I keep reading that I'm to "spin my fatigue" out. I kind of understand the whole lactic buildup and all but I have a hard time convincing myself to continue free spinning when my legs are jelly. So, assuming this is just a training (non-race) do I spin or relax on the way down?

2) I also keep seeing the term "recovery ride" in most training setups. I.E. the harder you ride one day, the less you're supposed to exert the next - or something like that. Let's say I go really hard for an hour (hills, intervals, etc..). Yes, I know an hour isn't a lot for most on here but I'm working my way up. And, if it means anything, by the end of the hour I can barely unclip due to fatigue. Do I really need to take a day off or do a recovery spin today? I want to get back on and go hard again - that's what she said - but I don't want it to be counterproductive.

Thanks in advance.

2) I also keep seeing the term "recovery ride" in most training setups. I.E. the harder you ride one day, the less you're supposed to exert the next - or something like that. Let's say I go really hard for an hour (hills, intervals, etc..). Yes, I know an hour isn't a lot for most on here but I'm working my way up. And, if it means anything, by the end of the hour I can barely unclip due to fatigue. Do I really need to take a day off or do a recovery spin today? I want to get back on and go hard again - that's what she said - but I don't want it to be counterproductive.

Thanks in advance.

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Using your leg muscles help in venous return of blood to your heart, so spinning your legs on the downhills help increases blood circulation and helps shuttle lactate from exhausted muscles to other places where it can be used.

Also if you push hard uphill then push hard downhill, that's twice the amount of training as when you coast down instead. A lot of people let up just as they crest a hill, so if you train to keep hammering up, over, and down, you can sometimes catch other riders unawares.

You don't get stronger when you're riding hard, otherwise your legs would feel great after pounding up a hill; you get stronger when your body is rebuilding during recovery from hard training efforts. If you always ride hard everyday, your efforts will eventually be mediocre. When you follow proper hard/easy modulation, you can train even harder on your hard days. Take a look at body builders; they never work on the same body parts two days on a row because they know they only improve when they give their body a chance to recover and rebuild itself stronger.

One of the biggest training errors endurance athletes make is to not make their easy days easy enough, so their hard days are not hard enough.

Also if you push hard uphill then push hard downhill, that's twice the amount of training as when you coast down instead. A lot of people let up just as they crest a hill, so if you train to keep hammering up, over, and down, you can sometimes catch other riders unawares.

You don't get stronger when you're riding hard, otherwise your legs would feel great after pounding up a hill; you get stronger when your body is rebuilding during recovery from hard training efforts. If you always ride hard everyday, your efforts will eventually be mediocre. When you follow proper hard/easy modulation, you can train even harder on your hard days. Take a look at body builders; they never work on the same body parts two days on a row because they know they only improve when they give their body a chance to recover and rebuild itself stronger.

One of the biggest training errors endurance athletes make is to not make their easy days easy enough, so their hard days are not hard enough.

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But that's for training only.

If you're riding competitively, trying to go as fast as your lungs and legs can drive you, it's generally faster to go hard on the climbs and coast on the descents.

It takes a LOT of power just to go a little faster on a steep descent - push 400W on a downhill and you might go from 45 to 47 mph in a descent that takes a minute or two and save 2-3 seconds. Ride up a climb that takes 10 min at 250W at 275W instead and you save a an entire minute.

That's assuming 1 min at 400W costs as much metabolically as 9 min at 275W, which are reasonable numbers.

All that's because the faster you go the more power you need to go even faster - power to overcome drag increases with the cube of speed. Doubling your speed requires 8 times the power to overcome drag.

And when you're descending a steep hill, you're getting a LOT of power from gravity. You can only add a relatively small amount of power for any decent length of time - even a smaller rider can get up to 3,000W of power or more from gravity on fast (50+ mph) descents. Since you're already going really fast on a descent, it takes a huge amount of power just to go a little faster.

If you're riding competitively, trying to go as fast as your lungs and legs can drive you, it's generally faster to go hard on the climbs and coast on the descents.

It takes a LOT of power just to go a little faster on a steep descent - push 400W on a downhill and you might go from 45 to 47 mph in a descent that takes a minute or two and save 2-3 seconds. Ride up a climb that takes 10 min at 250W at 275W instead and you save a an entire minute.

That's assuming 1 min at 400W costs as much metabolically as 9 min at 275W, which are reasonable numbers.

All that's because the faster you go the more power you need to go even faster - power to overcome drag increases with the cube of speed. Doubling your speed requires 8 times the power to overcome drag.

And when you're descending a steep hill, you're getting a LOT of power from gravity. You can only add a relatively small amount of power for any decent length of time - even a smaller rider can get up to 3,000W of power or more from gravity on fast (50+ mph) descents. Since you're already going really fast on a descent, it takes a huge amount of power just to go a little faster.

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playing word games.

if you truly exhaust your legs going up, it will not be possible to push going down. if you can push going down you didn't exhaust your legs going up.

if you truly exhaust your legs going up, it will not be possible to push going down. if you can push going down you didn't exhaust your legs going up.

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Free spinning (i.e. not fast enough for the hub to engage) I think is what the original poster is getting at. This takes minimal energy but accomplishes something.

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you are right, i misunderstood the OP. so it may make a difference.

*Last edited by hueyhoolihan; 01-01-15 at 08:11 PM.*

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I think it does help to spin out the lactic acid, it hurts right as you try and spin it out, but it reduces fatigue.

I don't always include an easy ride, but it really helps when you really start to train. This being you do most rides at a power that you have trouble holding.

I don't always include an easy ride, but it really helps when you really start to train. This being you do most rides at a power that you have trouble holding.

*Last edited by Bunyanderman; 01-02-15 at 12:28 PM.*

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...

I think this is pretty far off. As speed doubles, the power needed is 4x to overcome the air resistance. 3,000W would need to be a pretty steep climb with an ultra heavy object. On a 30% descent you experience half the acceleration of gravity (4.9m/s/s). If you weighed 70kg, this puts you at 343 newtons, or 343 watts. Unless you weighed 700kg, you would not experience 3000w of work.

I think this is pretty far off. As speed doubles, the power needed is 4x to overcome the air resistance. 3,000W would need to be a pretty steep climb with an ultra heavy object. On a 30% descent you experience half the acceleration of gravity (4.9m/s/s). If you weighed 70kg, this puts you at 343 newtons, or 343 watts. Unless you weighed 700kg, you would not experience 3000w of work.

72 kph = 20 m/sec

12% of 20 m/sec = 2.4 m/sec descent velocity

70 kg = 686N gravitational force

686N * 2.4 m/sec = 1646W since power is force times velocity

Bump that up to 90 kph on a 15% grade:

3.75m/s * 686N = 2572W

And that's just a 70 kg rider/bike riding normally. Bump that up to a 80 or 90 kg combination, or put the rider on aerobars or into a super-aero tuck and speeds will go way up as weight increases and/or drag decreases. And as descent speeds go up, power extracted from gravity also goes up.

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F = Cda * v**2

but power is

P = F * v

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I think this is pretty far off. As speed doubles, the power needed is 4x to overcome the air resistance. 3,000W would need to be a pretty steep climb with an ultra heavy object. On a 30% descent you experience half the acceleration of gravity (4.9m/s/s). If you weighed 70kg, this puts you at 343 newtons, or 343 watts. Unless you weighed 700kg, you would not experience 3000w of work.

How much power would you need to put out to go 50, 60, or even 70 mph on level ground?

Well, you're still overcoming that same amount of drag to go that fast down a hill.

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It may not be important to the question (or maybe it is?) but does spinning, jogging, or any similar low level exertion really flush lactic acid? Is flushing it out even desirable? It's something we've heard from way back, but I don't think it's true. Lactate is a fuel (although harder to metabolize), does not cause fatigue or soreness, and is not flushed faster with low level increased blood flow.

For example "this study suggests that blood flow may have an independent effect on net L at the upper extreme of the normal blood flow range during contractions but very little effect over a fairly wide low-to-middle range of flow rates."

I'm sure there are other good reasons to spin down the hill, and likely having to do with fatigue, so I'm not saying anyone should ignore the forgoing advice.

For example "this study suggests that blood flow may have an independent effect on net L at the upper extreme of the normal blood flow range during contractions but very little effect over a fairly wide low-to-middle range of flow rates."

I'm sure there are other good reasons to spin down the hill, and likely having to do with fatigue, so I'm not saying anyone should ignore the forgoing advice.

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This really depends on your level of fitness and training schedule. If you've had a hour long ride at a very high intensity, you can certainly ride the next day, but just ride at a lower pace. You shouldn't keep riding hard every day for weeks, since your body and muscle need so rest to rebuild. A typical month in my training include 3 weeks of building, followed by a full week of recovery. In an ideal week I'll ride 5 days/week.

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A 70 kg rider going down a 12% grade at 72 kph:

72 kph = 20 m/sec

12% of 20 m/sec = 2.4 m/sec descent velocity

70 kg = 686N gravitational force

686N * 2.4 m/sec = 1646W since power is force times velocity

Bump that up to 90 kph on a 15% grade:

3.75m/s * 686N = 2572W

And that's just a 70 kg rider/bike riding normally. Bump that up to a 80 or 90 kg combination, or put the rider on aerobars or into a super-aero tuck and speeds will go way up as weight increases and/or drag decreases. And as descent speeds go up, power extracted from gravity also goes up.

72 kph = 20 m/sec

12% of 20 m/sec = 2.4 m/sec descent velocity

70 kg = 686N gravitational force

686N * 2.4 m/sec = 1646W since power is force times velocity

Bump that up to 90 kph on a 15% grade:

3.75m/s * 686N = 2572W

And that's just a 70 kg rider/bike riding normally. Bump that up to a 80 or 90 kg combination, or put the rider on aerobars or into a super-aero tuck and speeds will go way up as weight increases and/or drag decreases. And as descent speeds go up, power extracted from gravity also goes up.

70 kg, 12%, 72 kph

70 kg = 686 N of force when perpendicular to the ground. The rider is not perpendicular, he is on a 12% slope.

To calculate this, (normal force) x (sin theta). 686 x (sin 12) = 140 N

140n x 2.4m/s = 336 W

90 kph, 15%

3.75m/s x [(686)x(sin 15)] = 675 W

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Seems like both of our physics games are off point.

70 kg, 12%, 72 kph

70 kg = 686 N of force when perpendicular to the ground. The rider is not perpendicular, he is on a 12% slope.

To calculate this, (normal force) x (sin theta). 686 x (sin 12) = 140 N

140n x 2.4m/s = 336 W

90 kph, 15%

3.75m/s x [(686)x(sin 15)] = 675 W

70 kg, 12%, 72 kph

70 kg = 686 N of force when perpendicular to the ground. The rider is not perpendicular, he is on a 12% slope.

To calculate this, (normal force) x (sin theta). 686 x (sin 12) = 140 N

140n x 2.4m/s = 336 W

90 kph, 15%

3.75m/s x [(686)x(sin 15)] = 675 W

675W is enough power to push the average 70 kg rider/bike combination to 90 kph on level ground?

The 1 km time trial track world record at sea level is almost exactly 1 minute - or 60 kph. World-class track cyclists can put out a helluva lot more than 675W for merely a minute - I've done more than that - and they're still slower than just about any random cyclist descending a 12% grade. Nevermind a 15% grade. And that world-class cyclist was using disk wheels, aero helmet, aero skin suit, shoe covers, and probably spent a lot of time in wind tunnels perfecting his positioning to reduce drag.

Your calculations are wrong.

Again - how much power would you have to put out on level ground to go 60 mph? That's how much power you need to get from somewhere to go 60 mph on a descent.

*Last edited by achoo; 01-02-15 at 11:09 AM.*

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Really?

675W is enough power to push the average 70 kg rider/bike combination to 90 kph on level ground?

The 1 km time trial track world record is almost exactly 1 minute - or 60 kph. World-class track cyclists can put out a helluva lot more than 675W for merely a minute - I've done more than that - and they're still slower than just about any random cyclist descending a 12% grade. Nevermind a 15% grade. And that world-class cyclist was using disk wheels, aero helmet, aero skin suit, shoe covers, and probably spent a lot of time in wind tunnels perfecting his positioning to reduce drag.

Your calculations are wrong.

Again - how much power would you have to put out on level ground to go 60 mph? That's how much power you need to get from somewhere to go 60 mph on a descent.

675W is enough power to push the average 70 kg rider/bike combination to 90 kph on level ground?

The 1 km time trial track world record is almost exactly 1 minute - or 60 kph. World-class track cyclists can put out a helluva lot more than 675W for merely a minute - I've done more than that - and they're still slower than just about any random cyclist descending a 12% grade. Nevermind a 15% grade. And that world-class cyclist was using disk wheels, aero helmet, aero skin suit, shoe covers, and probably spent a lot of time in wind tunnels perfecting his positioning to reduce drag.

Your calculations are wrong.

Again - how much power would you have to put out on level ground to go 60 mph? That's how much power you need to get from somewhere to go 60 mph on a descent.

Your mistake was using 686 N of force, this would only be true if you are perpendicular to the ground. Going down a 30% slope, you accelerate at half of gravity. At an angle of 90 degrees, you can calculate force of gravity on an object.

(normal force)x(sin 90) or normal force x 1

30%

(normal force)x(sin 30) or normal force x (1/2)

Were do my calculations go wrong?

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I never calculated the power needed to hit any speed. I only calculated the wattage that gravity applies to you at a certain speed, slope, and mass. I don't really know the power needed to hit 90 kph, but I would guess that it is more than 675W. Going 90 kph down a 15% slope and having the mass of 70 kg, gravity does 675 W of power. Nothing close to the 3000 W that was originally predicted.

Your mistake was using 686 N of force, this would only be true if you are perpendicular to the ground. Going down a 30% slope, you accelerate at half of gravity. At an angle of 90 degrees, you can calculate force of gravity on an object.

(normal force)x(sin 90) or normal force x 1

30%

(normal force)x(sin 30) or normal force x (1/2)

Were do my calculations go wrong?

Your mistake was using 686 N of force, this would only be true if you are perpendicular to the ground. Going down a 30% slope, you accelerate at half of gravity. At an angle of 90 degrees, you can calculate force of gravity on an object.

(normal force)x(sin 90) or normal force x 1

30%

(normal force)x(sin 30) or normal force x (1/2)

Were do my calculations go wrong?

First, the angle of a 12% grade is not 12 degrees (or radians! You get some fun answers if you use radians...). The precise angle is arctan( vertical rise / horizontal run ), which is arctan( 0.15 ) for a 15% grade, or 8.53 degrees.

Second, the normal force is m * g *

**cos**( theta )

Third, the normal force - the force perpendicular to motion - provides no power.

Fourth, if you're going to use the force parallel to the direction of motion, which is m * g *

**sin**( theta ), you need to use the speed of motion in the direction of motion. Not the component of motion in the vertical direction.

m is 70 kg, g is 9.8 m/sec**2, sin( 8.53 ) is 0.148, so the force from gravity parallel to the direction of motion is 101.8N.

And 90 kph is 25 m/sec.

25 m/sec * 101.8N = 2544W

That's a bit lower than my original 2572W, because in my first example I just approximated the vertical speed of the rider by applying the grade to the rider's speed along the road, which doesn't quite fit the exact definition of grade percentage as vertical rise divided by horizontal run because the length of the slope itself is longer than the horizontal run. But for smaller percentages it's awful close.

But if you want to figure out the rider's vertical speed, you'd just apply the sine of the grade angle to the rider's speed along the surface:

sin( 8.53 ) * 25 m/sec = 3.71 m/sec

Multiply that by the entire vertical force from gravity (because that's the vertical component of velocity), and we get:

9.8 m/sec**2 * 70 kg * 3.71 m/sec = 2544W.

Exactly the same.

It's a helluva lot easier to get power by approximating vertical speed directly from the grade percentage and multiply by the force applied by gravity.

25 m/sec * .15 = 3.75 m/sec

3.75 m/sec * 9.8 m/sec**2 * 70 kg = 2572W

Again, the power needed to go 90 kph down a grade is the same as the power to go 90 kph on level ground. You ain't going to go 90 kph on just 675W.

*Last edited by achoo; 01-02-15 at 11:54 AM.*

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Back to the OP, it depends. If you are riding the climbs at over LT, you'll build up more lactate than is useful for fuel. Lactate somewhere over 4 mmol (6 mmol?) will shut your muscles down. That has to be reduced to restore full muscle function. Therefore it's a good idea to spin easy, IMO enough to contract the muscles like you would on a recovery ride, thus stimulating blood flow. However, as has been pointed out above, lactate is also fuel, so if you haven't overloaded your legs, you can make better time by pedaling the descents, just not too hard. Ride length is a big determinate here, as eventually one runs out of glycogen. Remember that over LT efforts consume up to 10X the glycogen used in sweet spot efforts.

If you are trying to make your best time solo on a very long course, you won't be hammering the climbs. At the most you'll be climbing in your sweet spot. You should coast the descents to save your strength for the climbs. You'll be faster that way. No need to spin out the lactate, since it's useful for fuel.

Depending on conditioning, I think coasting becomes more favorable at over 60-100 mile distances.

If you are trying to make your best time solo on a very long course, you won't be hammering the climbs. At the most you'll be climbing in your sweet spot. You should coast the descents to save your strength for the climbs. You'll be faster that way. No need to spin out the lactate, since it's useful for fuel.

Depending on conditioning, I think coasting becomes more favorable at over 60-100 mile distances.

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What I think you're doing is shifting both the direction of motion and the direction of force. Shift one or the other to be parallel since power is the dot product of force and velocity vectors.

First, the angle of a 12% grade is not 12 degrees (or radians! You get some fun answers if you use radians...). The precise angle is arctan( vertical rise / horizontal run ), which is arctan( 0.15 ) for a 15% grade, or 8.53 degrees.

Second, the normal force is m * g *

Third, the normal force - the force perpendicular to motion - provides no power.

Fourth, if you're going to use the force parallel to the direction of motion, which is m * g *

m is 70 kg, g is 9.8 m/sec**2, sin( 8.53 ) is 0.148, so the force from gravity parallel to the direction of motion is 101.8N.

And 90 kph is 25 m/sec.

25 m/sec * 101.8N = 2544W

That's a bit lower than my original 2572W, because in my first example I just approximated the vertical speed of the rider by applying the grade to the rider's speed along the road, which doesn't quite fit the exact definition of grade percentage as vertical rise divided by horizontal run because the length of the slope itself is longer than the horizontal run. But for smaller percentages it's awful close.

But if you want to figure out the rider's vertical speed, you'd just apply the sine of the grade angle to the rider's speed along the surface:

sin( 8.53 ) * 25 m/sec = 3.71 m/sec

Multiply that by the entire vertical force from gravity (because that's the vertical component of velocity), and we get:

9.8 m/sec**2 * 70 kg * 3.71 m/sec = 2544W.

Exactly the same.

It's a helluva lot easier to get power by approximating vertical speed directly from the grade percentage and multiply by the force applied by gravity.

25 m/sec * .15 = 3.75 m/sec

3.75 m/sec * 9.8 m/sec**2 * 70 kg = 2572W

Again, the power needed to go 90 kph down a grade is the same as the power to go 90 kph on level ground. You ain't going to go 90 kph on just 675W.

First, the angle of a 12% grade is not 12 degrees (or radians! You get some fun answers if you use radians...). The precise angle is arctan( vertical rise / horizontal run ), which is arctan( 0.15 ) for a 15% grade, or 8.53 degrees.

Second, the normal force is m * g *

**cos**( theta )Third, the normal force - the force perpendicular to motion - provides no power.

Fourth, if you're going to use the force parallel to the direction of motion, which is m * g *

**sin**( theta ), you need to use the speed of motion in the direction of motion. Not the component of motion in the vertical direction.m is 70 kg, g is 9.8 m/sec**2, sin( 8.53 ) is 0.148, so the force from gravity parallel to the direction of motion is 101.8N.

And 90 kph is 25 m/sec.

25 m/sec * 101.8N = 2544W

That's a bit lower than my original 2572W, because in my first example I just approximated the vertical speed of the rider by applying the grade to the rider's speed along the road, which doesn't quite fit the exact definition of grade percentage as vertical rise divided by horizontal run because the length of the slope itself is longer than the horizontal run. But for smaller percentages it's awful close.

But if you want to figure out the rider's vertical speed, you'd just apply the sine of the grade angle to the rider's speed along the surface:

sin( 8.53 ) * 25 m/sec = 3.71 m/sec

Multiply that by the entire vertical force from gravity (because that's the vertical component of velocity), and we get:

9.8 m/sec**2 * 70 kg * 3.71 m/sec = 2544W.

Exactly the same.

It's a helluva lot easier to get power by approximating vertical speed directly from the grade percentage and multiply by the force applied by gravity.

25 m/sec * .15 = 3.75 m/sec

3.75 m/sec * 9.8 m/sec**2 * 70 kg = 2572W

Again, the power needed to go 90 kph down a grade is the same as the power to go 90 kph on level ground. You ain't going to go 90 kph on just 675W.

686 N of normal force and 8.53 angle at 90 kph gives you 2572 W yes you were correct and I was wrong, I do apologize.

But I think 90 kph is unreal, and 70 kph is more realistic.

686N x (sin 8.53) x (19.44m/s) = 1978W

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70 kg, 12%, 72 kph

70 kg = 686 N of force when perpendicular to the ground. The rider is not perpendicular, he is on a 12% slope.

To calculate this, (normal force) x (sin theta). 686 x (sin 12) = 140 N

140n x 2.4m/s = 336 W

90 kph, 15%

3.75m/s x [(686)x(sin 15)] = 675 W

edit: too late...

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2) I also keep seeing the term "recovery ride" in most training setups. I.E. the harder you ride one day, the less you're supposed to exert the next - or something like that. Let's say I go really hard for an hour (hills, intervals, etc..). Yes, I know an hour isn't a lot for most on here but I'm working my way up. And, if it means anything, by the end of the hour I can barely unclip due to fatigue. Do I really need to take a day off or do a recovery spin today? I want to get back on and go hard again - that's what she said - but I don't want it to be counterproductive.

Thanks in advance.

'Recovery' is necessary once in a while, not after every hard ride. If you're young, it might be enough to take off (or ride easy) one day a week. You can mix up your training but doing a classic 'recovery' ride after every hard effort is unnecessary.

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Also if you push hard uphill then push hard downhill, that's twice the amount of training as when you coast down instead. A lot of people let up just as they crest a hill, so if you train to keep hammering up, over, and down, you can sometimes catch other riders unawares.

You don't get stronger when you're riding hard, otherwise your legs would feel great after pounding up a hill; you get stronger when your body is rebuilding during recovery from hard training efforts. If you always ride hard everyday, your efforts will eventually be mediocre. When you follow proper hard/easy modulation, you can train even harder on your hard days. Take a look at body builders; they never work on the same body parts two days on a row because they know they only improve when they give their body a chance to recover and rebuild itself stronger.

One of the biggest training errors endurance athletes make is to not make their easy days easy enough, so their hard days are not hard enough.

think of it this way... do some biceps curls enough to trigger hypertrophy and generate lactic acid build up to a point where it burns and you have to stop... would you continue to do bicep curls at a lower intensity or would you stop and let the lactic acid dissipate before doing a next set?

do you see anyone benching pressing less weight in between their sets?