-5

There are numerous calculators online which estimate climbing speed based on power and grade or similar. For example:

Obviously, these calculators are only as valid as the input data. And since it is quite difficult to accurately measure CdA and Crr, the calculators really only produce accurate results on steep climbs where most of the energy goes into climbing. I'm not questioning that.

But, is it also important for the accuracy of these calculations that the climb is uniformly graded throughout? Or is it simply sufficient that the climb is steep enough throughout that aerodynamic drag never becomes too significant?

For example, Hawk Hill (a popular climb in the San Francisco area) averages 6% grade, but has sections as steep as 11% and some short sections that are flat, though none of them are long enough to build much speed. But I notice the calculators above lack any inputs that would capture this variability.

Will this significantly impact the accuracy of the calculation? Why or why not and how much? Please show with physics and math, if possible.

  • There's no way to apply an "average" to this -- every cyclist will be different. Consider that some cyclists would have no problem with an 11% grade while others would just about die on it, even though a 6% grade was was a piece of cake. – Daniel R Hicks Feb 18 '20 at 3:29
  • 2
    @DanielRHicks average grade has nothing to do with the cyclist. It's simply the difference in height divided by the difference in horizontal position. – Phil Frost Feb 18 '20 at 18:15
  • 1
    @DanielRHicks I fail to see how that's relevant. The definition of average grade is the same whether every cyclist is different, or if every snowflake is different, or if there's a teapot orbiting the moon. – Phil Frost Feb 18 '20 at 21:44
  • 1
    Google for R Chung method to determine drag and resistance. It will help you get accurate power estimates. – gschenk Feb 18 '20 at 23:37
  • 1
    @DanielRHicks or indeed the same cyclist on the same hill at the beginning/end of the ride, or having a off day. Or even 11% feels very different on a perfect road compared to having to ride over bumps and pick your way round potholes – Chris H Feb 19 '20 at 16:38
5

For the same average gradient and same constant power output, and assuming same bike, same tires, and zero wind the variable grade climb will take longer than a constant grade climb.

The reason is that with constant power and a variable gradient, you go faster when it's shallower (and slower when it's steeper) but aerodynamic drag increases with the square of air speed so you don't gain as much speed on the flatter parts as you lose on the steeper parts.

Suppose you were comparing two climbing routes: route A is a constant 5% over 3 km; route B is flat for 1.5 km, then climbs at 10% for 1.5 km. Both have a total length of 3 km, and both climb a total of 150 meters. Ignore for the moment the difference between road distance and horizontal distance.

At a constant 250 watts, the same CdA = 0.3 m^2, the same Crr = 0.005, rho = 1.2 kg/m^3, and total rider+bike mass of 80kg, the average speed for route A is 5.21 m/s, or a total time of 576 seconds.

For route B, the speed along the flat section is 10.51 m/s and the speed along the 10% gradient section is 2.98 m/s, for a total time of 646 seconds.

The above is just an extreme example but the same principles and calculations apply for any combination of gradients. That said, the smaller the range of grades around the average grade, the smaller is the difference between a climb at a constant average grade and a variable grade. For the example above where the average grade is 5% but half of the route was 0% and half at 10%, the difference in total time is 70 seconds. If, instead, half the route was at 4% and half at 6%, the difference in total time would be 3 seconds. This should not be surprising: the closer the variable grade is to a constant grade, the closer the two time estimates will be. That there is any difference at all is due to the nonlinearity of aerodynamic drag with speed, so any difference in speed for the shallower and steeper parts of the climb won't balance out.

Although you did not ask the question, constant power on route B is not the fastest way to do this climb. Constant power is time-minimizing only when the conditions are also constant so a time-minimizing strategy for variable gradient (or variable wind, or variable surfaces) is to vary the power. There are physiological constraints, of course, on how much you can vary the power so the optimization problem can be complex.

As an aside, I have spent some time examining the inverse question: given power and speed, can we calculate variable grade? Then I look at the calculated variable grade and find the drag parameters (CdA and Crr) that make the calculated grades match the actual road. This method of estimation appears to work well.

  • I think you may have missed this part of the question: "let's limit the issue to climbs which remain slow, steep, and sustained enough that still overcoming gravity is by a wide margin the most significant force for the rider to overcome." A "climb" where half the climb is flat is not that. – Phil Frost Feb 14 '20 at 0:05
  • 3
    I really like how you bring accurate numbers to these kinds of answers, and explain how they apply. I learn something pretty-much every single time. – Criggie Feb 14 '20 at 4:13
  • The question supposes that the differences in aerodynamic drag and rolling resistance are negligible. In such a case, does variable grade matter? – Phil Frost Feb 14 '20 at 21:40
  • 1
    Hmmm. Well, Crr is independent of slope so there's no difference in rolling resistance at all. However, as long as you're traveling within the atmosphere, aero drag will differ with speed. If you were traveling in a vacuum on the surface of the moon, there is no aero drag and speed would be proportional (linear with) gradient. In that case, variable grade does not matter and you could use average grade. On the surface of this planet, variable grade is slower than constant grade. – R. Chung Feb 14 '20 at 22:58
  • @R.Chung Crr may be independent of slope, but the faster you go, the bigger the rolling resistance becomes, so i'd have thought this would make variable climbs slower in a similar way to aerodynamic drag. trainingpeaks.com/blog/understanding-rolling-resistance – Andy P Feb 19 '20 at 10:46
2

If the grade did not matter, then you'd be able to maintain the same VAM on any grade. For climbs like you describe VAM is going to be the major determinant of overall time on the climb. For most riders their VAM vs grade forms a bell curve. Based on your fitness and gearing, there will be a grade on which you can maintain your maximum VAM. For shallow grades, you simply can't ride fast enough to maintain the cooresponding vertical velocity and for steeper grades you are limited by the gearing to cadences that limit power output.

The wider your gear range the steeper grade you'll be able to sustain your maximum VAM. However, eventually even the lowest gear will require more power than you have to keep the same VAM.

For an interesting application of this see this article on the 48 hour climbing record. The rider chose a very specific section of climb that had a gradient on which he could record his personal maximum VAM. I've ridden the entire climb many times, I can't imagine just doing the steep part for 48 hours.

Now as far as the meat of your question goes, I'd say it's depends on how you're using the calculator. For relatively steep climbs at slower speeds, the weight of the rider/bike is one of the most important inputs. The time on the climb is linearly related to the weight (i.e. 1% error in weight estimates == 1% error in estimated time).

If your using them to get some estimate of your avg power on a climb, then using the average gradient is an acceptable estimate of the total work as long as gradients are within the range of your gearing. If you are using them to estimate your time on a climb, then the inaccuracy of the input data far outweighs any variance due to gradient change. (i.e. your estimates of bike/rider weight and avg power likely far outweigh any errors in the formula caused by changing grades in the climb).

  • "then the inaccuracy of the input data far outweighs any variance due to gradient change" in accuracy of what input data exactly? Why would the calculations be valid in one direction (calculate time by average power) but not in the other (calculate average power by time)? – Phil Frost Feb 14 '20 at 21:42
  • 1
    @PhilFrost Actually, if you know gradient, CdA, Crr, the air density rho, and the total mass then, yes, you can accurately calculate both average power if given speed, and speed if given power. The exact equation is well-understood if not always well-known. You can find a discussion of it here in this bikes.SE question and answer. – R. Chung Feb 15 '20 at 4:54
  • It is one thing to put an power output into a form, it is completely another to maintain that power output steady for an entire climb. Is your water bottle full? What is your exact weight at the bottom of the climb? What does your bike really weigh? – Fred the Magic Wonder Dog Feb 16 '20 at 0:01
  • What I meant is that after the fact, you can make pretty good estimates of how much you weighed on that day and what your bike weighed. Since for steep climbs there is a roughly linear relationship btw weight and time, even minor % errors in estimates show up. – Fred the Magic Wonder Dog Feb 18 '20 at 16:42
  • @FredtheMagicWonderDog OK that makes sense. I was confused because I took the wording of the last sentence in your answer to imply (due to it's juxtaposition with the previous sentence) input data inaccuracy somehow makes the calculation uselessly inaccurate. – Phil Frost Feb 18 '20 at 21:48
0

In practical terms, it is not possible to rule out the effects of aerodynamic and rolling resistance.

Bike calculator actually has an old tool that would help with this problem. It gives the option use up to 6 segments. http://bikecalculator.com/tripUS.html

I spent a bit of time with strava to break the hawk hill climb into segments (distance-elevation gain)

 .22mi - 149ft
 .88mi - 134ft
 .56mi - 223ft

Firstly I entered sample data for myself (130lbs, 20lb bike) on a single segment representing hawk hill: 1.66mi - 506ft.

At 275W, this gave an estimated climb time of 8.25mins

Next I used the 3 segments - to produce the same time (8.24mins) required 289W. Not a terrible estimation. Certainly good enough to prove if a power meter is 25% out.

However it is easy to use the tool to show that on this climb at least, aerodynamics are still going to cause a significant problem. If I add a 3mph head wind (hard to gauge accurately in real life), and change position from hoods to bartops (to represent a worse than average CdA) then 318W are now needed.

When comparing these figures to the 'rider in a vacuum' scenario, it predicts the following:

68 x 9.8 x 154 = 102626J

102626 / 275 = 373s = 6.22 minutes

Bike Calculator predicts that to achieve a time of 6.22mins in real life I would need to produce 475W rather than 275W - this ties very well with the recorded power values on Strava. So for the provided sample climb, the rider in a vacuum example underestimates power by approximately 73%

-2

Provided the basic assumptions behind the calculator (that aerodynamic drag and rolling resistance are relatively small) are maintained, the calculator remains accurate even if the grade is somewhat variable. This can be shown by considering just the gravitational part of the equation, which depends only on the difference in height and the total mass.

Climbing higher requires increasing gravitational energy:

gravitational energy = mass * gravity * height

On Earth's surface, gravity is 9.8 meters per second-squared.

So for example, if the bike and rider weigh 90 kilograms, and wish to climb 100 meters, this requires an increase in gravitational energy of

90 kg * 9.8 m/s^2 * 100 m = 88200 joules

(check it on Wolfram Alpha)

A watt is one joule per second. So riding at a power of 200 watts, this climb will take:

88200 joules / 200 watts = 441 seconds

Or 7 minutes and 21 seconds, in a frictionless vacuum where the rider begins and ends with the same kinetic energy (speed).

It does not matter for this calculation if the grade is 1% or 20%, or variable throughout the climb. It doesn't even matter if there are some downhill sections, since the kinetic energy gained on the downhill sections will be carried into the uphill sections.

So, a variable grade makes no difference to calculating power required to overcome gravity: only the height difference matters.

However, aerodynamic drag is proportional to the square of velocity. The calculators estimate the power to overcome this drag by assuming a constant speed. However riding some sections at a slower speed and others at a higher speed requires more energy than riding at a constant speed over the same distance in the same time.

Thus, if the climb is ridden at a constant effort rather than a constant speed, the calculation becomes less accurate. But as the accuracy of the calculation already depends on aerodynamic drag being a small component of the energy, this has only negligible effect on the accuracy of the calculation.

  • 1
    Was not me, but the question specifies "real world climbs" with variable grades, and AFAIK there aren't a lot of frictionless vacuums out there. There are a number of time estimators as listed in your question, and they have their place in direct comparisons with themselves, but again are idealised. Perhaps the best estimator is the strava log for all your own personal efforts on that particular climb ? – Criggie Feb 18 '20 at 23:19
  • 2
    You don't have to go such lengths of you had first year physics. All line integrals in conservative vector fields are independent of the choice of path. Now you have to only ensure you are dealing with a conservative field in good approximation. Firstly, cycle at non-relativistic speeds. Second, be large enough to remain in the regime of classical physics. Third, go slow to make velocity dependent friction small. Lastly, have difference in steepness small enough that overall track length difference can be neglected (friction). – gschenk Feb 18 '20 at 23:35
  • 3
    Right, I think the problem is the discord between the question ("real-world climbs") and the answer ("frictionless vacuum"), and what constitutes amounts that are "negligible" to the OP. – R. Chung Feb 19 '20 at 0:10
  • 1
    Down voted as I find it hard to breathe when climbing hills in a vacuum. Also I find frictionless tyres to be somewhat tricky to ride on. I did however upvote the question itself as it's an interesting topic – Andy P Feb 19 '20 at 10:32
  • 1
    @AndyP it is not the point of an approximation to re-create the conditions of it (i.e. ride in vacuum). But to reduce the solution to what is solvable. Here we are looking not at the absolute friction, but the difference in overal friction between two paths. For not very large difference in gradients the difference in path lengths is not large. A 8% 1km ramp will be shorter than a route that undulates between 6% and 10% and averages to 8%, but not much (apply x = sin x approx). The average gives a lower bound, and a good approximation. Thus we may not need to consider friction. – gschenk Feb 19 '20 at 12:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.