# Tech FAQ: Readers answer the bike weight question

## Lennard Zinn defends his stance on bike weight from previous columns

*Editor’s Note:** Lennard Zinn’s regular column is devoted to addressing readers’ technical questions about bikes, their care and how we as riders can use them as comfortably and efficiently as possible. Readers can send brief technical questions directly to Zinn*.

**Dear Lennard,**

One comment on this article on the importance of bike weight.

Let me start with a (not very funny) physics joke: A farmer asks a physicist how many cows he can fit in his barn. The physicist goes off to perform some calculations and returns the next day. The physicist begins his explanation: “Let’s first assume the cow is a sphere…”

Here, Miguel’s calculation is flawed because the assumption is the bike only goes one direction: up an incline at a constant speed. Fine, except that’s not even remotely true. His calculation is probably right (I didn’t check the math) as far as its assumptions, but no one rides this way, especially climbing. The bike goes all over the place, including rocking and weaving. Light bikes “feel” lighter in my opinion in large part because of all the non-linear, forward movement. Heaving 29 pounds back and forth up a hill is going to wear you out much more than heaving 16 pounds back and forth. There’s a reason every cyclist thinks weight matters. I’m not going to try to prove this, but I would propose this experiment: take each bike one at a time and ride standing while waggling your bike back and forth (on a flat road). I guarantee they will feel much different.

Needless to say, same for rotating wheel weights: weight wouldn’t matter so much if your rpms were constant, but most people accelerate/decelerate with every pedal stroke, especially when climbing. It’s the constant acceleration/deceleration that’s going to get you.

— *Mike*

**Dear Mike,**

Actually, in a frictionless universe, that acceleration and deceleration due to varying pedaling efforts makes no difference, because you get the energy back. The bike keeps going when you stop pedaling, and you benefit from the bike’s momentum and the wheels acting like lightweight flywheels. It is only when you apply the brakes (i.e., you give up some of your hard-earned energy — or work done — as heat) that there is a net energy cost to accelerating the bike back up to your original speed.

We don’t live in a frictionless universe, but relative to the kind of power required to propel a 180-pound bike and rider up a mountain at 12 mph, drivetrain friction and rolling friction can probably be ignored. The biggest variable would be wind, but on a calm day, or especially with a steady tailwind similar in speed to rider speed, I think that the frictionless universe analogy used in elementary physics calculations of motion is not a bad one.

Which brings us to your comment that “the bike goes all over the place, including rocking and weaving.” Weaving, if we are able to ignore rolling friction, including tire scrub as the wheels veer back and forth, will be irrelevant as well. What matters for the calculation is the net altitude gained, and if you effectively decrease the grade by weaving, that costs you nothing, too.

As for rocking the bike, yes, it takes more energy to lift a leaned-over bike, but you also get that back when leaning it the other way, and when the lean angle is quite small, which I think it is for most people, I doubt it’s relevant, either.

— *Lennard*

## Readers sent in a number of other takes on bike weight from my recent column. Here are a few:

**Dear Lennard,**

Another way to look at this question would be from a potential energy perspective. Potential energy from gravity is equal to Mass x Gravity x Height. For a given climb, Mass and Height will be the same; the weight of the bike would be the only variable.

Assume an 80kg rider and a climb with a 100-meter elevation difference.

Joule is the unit of potential energy when using SI units.

**Bike 1:** 93.18 Kg x 9.8 m/sec^2 x 100 m = 91,316 joules

**Bike 2:** 87.72 x 9.8 x 100 = 85,965 joules

The difference is 5,350 joules.

Now assume that the climb took the same amount of time on both bikes (500 seconds).

A Watt is equal to a Joule/second

The heavier bike would require an additional 10.7 Watts of effort.

Another way to illustrate the difference would be to assume that the rider effort (Watts) is constant on either bike.

Assume 175 Watts power output.

**Bike 1:** 91,316 joules / 175 Watts = 522 seconds

**Bike 2:** 85,965 joules / 175 Watts = 491 seconds

The time difference is 5.8 percent (very similar to your calculations).

— *Mike*

**Dear Lennard,**

Miguel’s calculation only looks at force, and ignores power. Certainly you can crawl up the slope at 5 mph, but when you’re trying to climb at 15 mph, the power difference is more marked. (From the diagram, I calculated that the drag from a 12-pound weight difference is about 10 N. At 5 mph, or 2.2 m/s, the power difference is 32 W. At 15 mph, or 6.7 m/s, the power difference is 67 W.)

Furthermore, power itself isn’t the whole picture, as the (human) motor is highly non-linear. You can’t increase power indefinitely; at some point you’ll reach a threshold and will not be able to produce additional power. Furthermore, everyone has different capacities to do work, i.e. power over time. One cyclist may be able to do 250 W over an hour until exhaustion, while another may only be able to sustain that pace for 45 minutes. That’s why I don’t quite buy the extrapolated or interpolated figures from the studies you cite.

*— Lawrence*

**Dear Lennard,**

I love the discussion about climbing and weight of the bike. Here’s my $.02: I would be skeptical about any numbers comparing performance with weight of various bikes if the weight of the rider is not included. I buy the experimental data you presented about the ride up l’Alpe, but you can’t expect exactly the same results for every rider.

It’s the total weight being dragged up the hill that determines how much work is being done. The bike’s fraction of the total weight is much greater for lighter people and less for heavier folks. This makes bike weight a bigger issue for the lighter rider and less for the big guys.

Hmmm… If I’m 20 pounds overweight, I can keep my steel frame in good conscience. If I just spent $10,000 on a featherweight, I’m throwing money away unless I jettison the gut!

*— Michael*

**Dear Lennard,**

One thing that I think was overlooked in your “Does bike weight matter?” response was the idea of “overall” weight (rider + bike and equipment). While the difference between a 17-pound carbon bike and a 29-pound steel tank is significant, it’s less significant when you add in the weight of the rider. Moreover, as the size of the rider increases, the weight difference of the bike becomes even less relevant. So while a 29-pound bike is 70 percent heavier than a 17-pound bike, if you throw a 160-pound rider into the mix, the 189-pound combo is only seven percent heavier than the 177-pound combo (same rider, 17-pound bike). Obviously, if the rider were 200 pounds, the percentage difference would decrease further due to the bike’s weight being proportionally lower to the total weight.

So while there’s obviously a fairly big difference between a 29-pound bike and a 17-pound bike (12 pounds if my math is correct), if somebody is thinking of spending thousands to “upgrade” from an 18-pound bike to a UCI-limit 15-pound bike, their expectations of vast improvements on climbs should be tempered.

— *Steve*

**Dear Lennard,**

Referring to Miguel’s question today, isn’t the “elephant in the room” (so to speak) the weight of the rider, and indeed, the total loaded weight of bike and rider? True, on a 20-percent slope the down force of 17 pounds is about 3.3 pounds, and 29 pounds about 5.6 pounds, as Miguel’s drawing shows. If the rider weighed zero (just to pick the extreme case), that would require nearly twice the power to ascend at the same speed. If the rider weighed 350 pounds, the weight differences of the bikes would mean zilch (except for structural reasons). But with a relatively light rider, bike weight is clearly a factor.

If the weight of everything else (rider, clothes, shoes, water, patch kit, etc.) is, say, 170 pounds, and the bikes being compared weigh 17 or 29 pounds, then the true comparison is between 187 and 199 pounds, or down force of about 36.6 vs. 39 pounds, or six percent. Actually, you don’t even need to do the trig calculation of down force, because the answer is the same based simply on a comparison of gross total weight — 199 is about six percent more than 187. Of course, aero is a factor too (at higher speeds), and the weight helps downhill, but the downhill gain is precious little compared to schlepping it uphill. You can quibble about rolling resistance, drivetrain drag, rotational weight, and aero, but weight is a big deal — both for the bike and the rider.

— *Ron*

**Dear Ron and Steve and Michael,**

You all make very good points, and I want to clarify a couple of things. The added power required to pedal each additional pound of weight on the bike or rider up the hill at a given speed will not change with the total weight of the bike and rider. However, the percentage power increase required to do it will decrease as the total weight of the bike and rider increase. And that’s where Lawrence’s comments about the human motor being non-linear come into play. A 100-pound rider is generally less able to make, say, a 50-Watt power increase than is a 200-pound rider.

— *Lennard*