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Bike components or upgrades are sometimes quoted as saving some number of "grams" of drag.

How can a gram, a unit of mass, be used to quantify drag, which is a force?

And for extra credit, why is this a useful way to quantify drag?

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  • 5
    One view is on a set of scales, a gram is easy to measure, easy to compare and easy to charge a fortune if you can shave a few off. Converting drag to grams makes it easy to charge suckers more for something very few really understand and even fewer benefit from.
    – mattnz
    Jan 22, 2020 at 0:06
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    It doesn't make sense as a force either. Or do they specify a speed?
    – Nobody
    Jan 22, 2020 at 12:04
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    @Nobody Sometimes. But that's a different matter.
    – Phil Frost
    Jan 22, 2020 at 14:47
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    You can't make the Kessel Run on your bike in under 12 parsecs without reducing your grams of drags...
    – Michael
    Jan 23, 2020 at 19:24
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    @chiliNUT I would argue that it makes infinitely more sense for mechanical keyboards. They (in theory) have the same amount of force to activate regardless of how quickly you press them. You need to overcome the spring force to actuate them enough to register, and since spring force only depends on travel distance, they can assume the keypress has a set amount of force to activate. Converting mass to equivalent weight force isn't really the problem. The problem is that the drag force will depend completely on your travel speed, unlike a keyboard press.
    – JMac
    Jan 23, 2020 at 19:40

4 Answers 4

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You're exactly right, drag should be measured in units of force, like Newtons. However, although there are small differences in the gravitational constant across the surface of the Earth, a reasonable rule of thumb is to assume it to be constant. Accordingly, drag force in Newtons is proportional to kilograms by a constant term very close to 9.8 m/sec^2, so 1 N of drag force is approximately equivalent to "100 grams of drag."

That said, I try not to use "grams of drag" and instead use Newtons of force. But, as you've undoubtedly seen, many others (including bike manufacturers and wind tunnel operators) do use "grams" so in order to promote effective communication, I'll switch to using a term I know they'll understand.

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    It needs to be noted that "pound" is both a unit of mass and a unit of force. And Merriam-Websters says that "gram" can be a unit of force as well. Jan 22, 2020 at 0:28
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    Might it be used because the user can hold a 1 gram mass and perceivable its weight, and can visualise how much the air is holding back the bike ? Understanding a weight in grams or pounds due to gravity of an item in your hands is more relatable than doing the same in Newtons , though functionally the same ?
    – Criggie
    Jan 22, 2020 at 2:51
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    @Criggie Yes, and that's why engineering is littered with way too many standards. Because everyone thinks something else is more sensible.
    – Mast
    Jan 22, 2020 at 9:02
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    The "pound" is in usual practice referring to the pounds-force. There is a distinct unit for pounds-mass. At "standard temp and and pressure" they are equal. engineering.stackexchange.com/questions/2300/… Jan 22, 2020 at 18:07
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    @AndrewMorton Chris Yu from Specialized discussed how hard it is to explain aero drag in this podcast: cyclingtips.com/2019/07/… and Josh Poertner of Silca discussed how hard it is to explain rolling resistance drag in this podcast: marginalgainspodcast.cc/… In both cases, I think there's a move toward using CdA and Crr to characterize drag since they're independent of speed (at the speeds cyclists typically ride).
    – R. Chung
    Jan 25, 2020 at 2:00
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How can a gram, a unit of mass, be used to quantify drag, which is a force?

It's a misnomer related to the Kilogram-force (kgf) and its submultiple unit the gram-force (gf), 1 kgf being equal to 1000 gf.

A kilogram-force is equal to the magnitude of the force exerted on 1 kg of mass in a gravitational field of 9.80665 m/s² (standard earth gravity), it is by definition equal to exactly 9.80665 N.

This unit is only useful to communicate with people unfamiliar with the Newton unit. The kilogram-force (and its cousin the Pound-force (lbf)) is easier to grasp for the interested layman.

It is however considered an unacceptable unit in the NIST recommendations and is often replaced by the decanewton (dN) in technical documentations, 1 dN being equal to 1.01971621 kgf.

A gram-force is equal by definition to 9.80665 mN (but often rounded to 10 mN for marketing purpose), so when one say "1 gram of drag", it actually means "1 gram-force of drag" which usually mean "10 mN of drag".

Please also note that the drag force is proportional to the square of the speed so any mention of drag force should specify at which speed otherwise it's just marketing claptrap.

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  • Is gram-force derived from kilogram-force or is it a survivor from CGS units?
    – houninym
    Jan 22, 2020 at 10:40
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    @houninym I don't know but I don't believe gram-force was widely used before CGS was superseded by MKS and SI. Either way, I did not used derived in the historical sense, so I changed it to submultiple to avoid confusion.
    – zakinster
    Jan 22, 2020 at 11:17
  • Should we credit this answer to wiki ?
    – Dan K
    Jan 22, 2020 at 12:02
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    @houninym No, the Dyne and the gram-force is not the same thing. The Dyne is defined by the force to accelerate a mass at a defined speed (same definition that the Newton but with CGS units instead of MKS). The gram-force is however defined by the gravitational acceleration. A Dyne is to CGS what the Newton is to MKS and is by definition equal to 10^-5 N. 1 gram-force = 9.80665 10^-3 N = 980.665 dyne.
    – zakinster
    Jan 22, 2020 at 16:24
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    @sjakubowski do you have an authoritative source for that ? Because since the drag at 40km/h would be twice the drag at 28km/h, I wouldn't be suprise if manufacturer advertise drag at lower speed unless there's a law/regulation against that.
    – zakinster
    Jan 27, 2020 at 8:54
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How can a gram, a unit of mass, be used to quantify drag, which is a force?

Only by using a gram as a shorthand, or malapropism depending on your perspective, for a gram-force, which is the force exerted on a gram mass by (Earth's) gravity.

why is this a useful way to quantify drag?

In the context of cycling this is simple: it allows the main energy costs that cyclists often seek to reduce to be handled with the same units.

As well as reducing drag, cyclists are also often concerned with reducing mass. The main value of reducing mass is that it reduces the amount of additional power from the rider required to climb uphill (it also reduces the work needed to accelerate, but most riders in most scenarios end up putting more effort into climbs).

This additional power is used to work against the gravitational force pulling the rider and bike down (and consequently back downhill). Reducing mass thus reduces weight. It is really these two forces – drag and weight – that cyclists are usually most interested in reducing.

We could choose to measure them both in a unit of force (Newtons, say) and this would speak more directly to the value of reducing them, but the reality is that it's easier to pop a derailleur on a set of kitchen scales and get a readout in units of grams of mass (though even the scales are really measuring a weight force) than it is to fire up a wind tunnel and get a balance readout in Newtons of drag.

The value of a Newton of weight compared to a Newton of drag isn't a 1:1 correspondence but because the physics that makes the forces important in cycling overlaps a lot, dealing with both in the same units does help with developing an intuitive sense of one when you've already learned a bit about the other. A gram is probably as good a compromise choice for this as any other.

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  • The advantage of using the same unit name for force and mass is that the average person can better relate the measurement to reality. Pretty much everyone has a "feel" for how much force a pound or kilogram weight produces -- just lift one and you know. Jan 22, 2020 at 14:13
  • But a gram of drag which pulls you backwards is not at all the same of a gram of mass unless you're riding up a vertical cliff, right?
    – Phil Frost
    Jan 22, 2020 at 14:44
  • @PhilFrost correct, that's the thrust of my last paragraph. It just removes an additional conversion you need to do to compare the two. If you measure drag in Newtons and mass in grams, you have to know that 100g weighs 1N and what the sine of the slope is.
    – Will
    Jan 22, 2020 at 17:00
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It doesn't make any sense whatsoever to try to substitute a constant amount of weight for a force that is a function of speed.

Obviously the calculation is done by choosing a value for the speed, evaluating the drag force and dividing it by the gravitational acceleration g.

This doesn't provide any insight into the actual benefit of a reduction of drag. The only purpose is to be able to brag about it or sell it to people who don't understand the concept.

To understand how wrong it is, consider that people's average speeds easily vary by a factor of 2, so the actual drag varies by a factor of 4. If you compare climb speeds with flats, speed might even vary by a factor of 4-5 for a drag variation by a factor of 16-25. So your "easily understandable" value is wrong by a whole order of magnitude if you don't know how exactly to use it. Another way to be off by orders of magnitude is because for real weights you lift them up by the vertical climb amount which is rarely more than single digit kilometers, while for pretend weights you "lift" them forward by distances that can easily reach triple digit kilometers.

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    Wind tunnel tests for bike components are typically done at a single speed (which you can find in the footnotes or appendix of any report or study of drag). That said, when bike companies state the "grams of drag" for different configurations they're comparing at the same speed. All you have to do is read the fine print to figure out what speed that is.
    – R. Chung
    Jan 24, 2020 at 16:24
  • @R.Chung And if you then want to compare different tests/manufacturers with your use case you need to do calculations which are certainly beyond people who couldn't handle a drag coefficient. On the other hand, people who couldn't do those calculations would still be able to compare drag coefficients because that doesn't involve any calculation (other than evaluating greater/smaller than operations).
    – Nobody
    Jan 24, 2020 at 22:08
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    Yeah, we discussed that in this podcast: cyclingtips.com/2019/07/… I prefer to drag in terms of CdA and Crr (since they're speed-independent at the speeds that cyclists typically ride at) but usually I get blank looks. In that podcast I think Chris Yu talked about how hard it is to explain drag in ways that the public can understand.
    – R. Chung
    Jan 25, 2020 at 1:57

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