Drag to time savings

In the two following sheets you can find different models to translate drag to time savings. The first model is the most complex and it leaves the door open for a pretty unintuitive result, small drag savings measured in the wind tunnel could translate to time losses in real world. I have tried to find out when this could happen but, for the moment, I haven't found any real world combination of parameters that could produce time losses.

To compare the performance of the 3 models that can be used for real world conditions (non-null wind speed), I have considered two different bike speeds for the initial run (R1). For both bike speeds, wind direction and speed have been chosen to produce a yaw equal to -10º for the initial run. Then, a CdA reduction equivalent to 50gF measured in the wind tunnel at 30mph has been applied and time savings are calculated for each model. For the model 1, I have considered Speed Concept drag-yaw linear relation (DZ on, Fastest bottle configuration) taken from Cervélo P4 in the Tunnel Slowtwitch's article. Yaw angle is considered negative when the apparent wind speed sees the driveside first. Here are the results:

As you can see, the phenomenon that causes the paradox that I have already commented is present here. Drag savings in real world are smaller than in the wind tunnel because the speed increase causes a yaw reduction and, consequently, a small drag increase. The error is maximized for yaws near stall angle, when the derivative of drag with respect to yaw is bigger.

The previous picture shows the formula that's typically used to convert from drag to time savings. As you can see, it's pretty accurate for most situations, especially when bike speed is similar to wind tunnel speed. Drag-time relation is pretty linear for all the models that I've presented so it's also accurate to predict time savings for bigger drag reductions.

That's all for today. Thanks for reading!

Drag to time savings. An introduction

As an introduction to this topic, two plots that show the relation between time savings, CdA reduction and previous CdA for a given bike speed. These two plots are obtained for the only situation where time savings are independent of wind speed and direction, the null wind speed case. The first plot shows time savings when the variations of friction and gravity power due to the speed increase are neglected. The second plot also considers these variations

As you can see, when the variations of power due to gravity and friction are neglected, time savings are overestimated. In the following posts, I will present more complex models that consider wind speed, direction and the modification of yaw angle and, consequently, CdA.

More soon. Thanks for reading!

A tool to determine airfoil shape using rim cross section

Just a small tool to see how the airfoil shape of a wheel is modified when the distance from the ground to an horizontal cut plane is changed. As the wind is constrained to flow parallel to the ground, this airfoil shape is what the wind sees depending on the distance to the ground. Bontrager commented some time ago that they tried to improve the performance of their airfoil shape for h=0 (h is the distance from the hub axle to the horizontal cut plane) because the aspect ratio is the lowest and this was the worst case scenario. I haven't got neither CFD nor wind tunnel data to back this up but, as you will see in the following graphics, chord length and the shape of the airfoil changes notably when h is modified. To illustrate how this tool works, I will use the following rim cross section:

NACA 0030 (3.333:1) symmetric airfoil with an overlaid 20mm tire. This could be assimilated to a 70-75mm rim with tall blades and curved brake walls. 27mm at widest point

The following plot shows the tire leading side for some values of h:

Wheel radius, 0.34m. Yellow, h=0.1m. Green, h=0.21m. Blue, h=0.26m. Black, h=0.32m. As you can see, when h increases, the widest point approaches to the leading edge of the airfoil. When h>wheel radius-chord length, tire leading and rim leading sides form an unique airfoil

Finally, in the following graphic, the relation between aspect ratio and h for the interval with separated tire leading and rim leading sides airfoils is shown :

It goes from 3.333:1 to 8.535:1. Average is 4.069:1

That's all for today. Greetings

A yaw-independent quantification of aerodynamic performance

The relation between yaw and aerodynamic drag makes it difficult to discern which component is really faster. Wheel manufacturers usually use sentences like "X wheel is Y seconds faster than Z wheel  over 40km" but how do they calculate that time gain if wind and bike speed is variable as well as drag as a function of yaw? The answer is using a weighting function that multiplies the drag function obtained in the wind tunnel by a factor to obtain an unique drag value.

Until now, the weighting functions that every manufacturer use haven't been known but Mavic showed his during the presentation of the Mavic CXR80. In the previous post, I worked that CDF function to obtain a normal distribution that can be used as a weighting function. Next step is to consider the drag function obtained in the wind tunnel. Typically, aerodynamic drag is considered to vary linearly between measurement points. Another aspect that has to be considered is that aparent wind speed has been variable for that weighting law but, as the fluctuations of the Reynolds numbers are small, CdA can be considered independent for a given yaw.

Let's consider a drag function with n divisions, the first abscise is called x0 and the last one, xn. The CdA for every abscise is called di, with i going from 0 to n. The weighting fuction is assumed to be a null-mean normal distribution as Mavic has showed. Now, we can define the weighed yaw-independent CdA as:

erf, error function. sigma, standard deviation of the weighting function. The leading term (before the summation) takes into account that the drag function isn't defined in the whole range of possible yaw values

Let's make an example. Consider the following drag function taken from Slowtwitch's Cervélo P4 in the Tunnel article:

As you can see, the Cervélo P4 and the Trek Speed Concept are nearly tied. Considering that wind tunnel speed was set to 30mph, approximated CdA values are:

Using Mavic's weighting function with a standard deviation of 10.3388º, the terms of the summation over each interval are the following ones:

The leading term is equal to 1.056, the closer this term is to the unit, the surer we are of the conclusion. Finally, the yaw-independent CdA values of both bikes are:

The conclusion is: "A riderless Cervélo P4 "ridden" in typical conditions will produce 100*(0.003868-0.039685)/0.03868=2.598% less drag than a Trek Speed Concept with a confidence of the 100/1.056=94.697%". As the correlation between wind tunnel data and real world data has been already proved, this method, when coupled with pedaling-rider wind tunnel data, can answer the question "Which setup is fastest?" without any doubt. 

That's all for today. Thanks for reading!

About yaw

The subject of yaw is of vital importance when trying to design aerodynamic components. Typical values of yaw seen during cycling are intimately relationed with the speed of the cyclist and, therefore, the target client of these components

Dan Connelly developed some time ago a probability function for yaw motivated by some conflicts between ZIPP and HED about typical yaw values. This probability function used a Maxwellian distribution to describe wind speed and assumed that all the headings and wind directions were equally probable. The relation between wind speed at the height of a meteorological station and at bike level was obtained using the Hellman formula. I haven't checked all the process to obtain this function but Dan is a clever guy so I don't expect any mistakes to be found. My only concern was the calculation of v0, the standard deviation of the velocity components of the Maxwellian distribution. Dan assumed, maybe using a conclusion that is valid for normal distributions, that the peak of the probability function was equal to the average value of the probability function for speeds. Hellman formula works for average values of wind speeds so a correct relation between v0 and the average wind speed should be used. For this Maxwellian distribution for speeds, the average is given by the following limit:

 Working it, we get:

Now we can calculate v0 as a function of the average wind speed at bike level that is obtained using the Hellman formula. The following two graphics show the influence of this modification in the wind speed and yaw probability functions that Dan showed.

As you can see, considering that the average wind speed is equal to v0, there is an underestimation of the presence of low wind speeds and yaw values.

Mavic showed a probability function of yaw determined through experimental studies during the development of the new CXR80 wheel. There isn't many details about how that function was obtained (wind speeds, bike speeds, etc) but I have tried to replicate it using the yaw probabilty function that Dan developed. First, I fitted a normal distribution to Mavic's plot. The plot misleads a little bit because it shows the CDF between every tick - 1.25º and + 1.25º but I have calculated the standard deviation using the CDF of an arbitrary division of the x scale. The normal distribution and the CDF of yaw obtained from Mavic's data are the following ones:

Explanation: If you draw a vertical line taking as origin any abscise (let's call it Y), the ordinate of the intersection between that line and the function multiplied by 200 represents the percentage of time that you ride in the interval [-Y,Y]

That's all for today. Greetings

An integral approach to rolling resistance III

In the previous post of the series, I considered a simplificated relation between contact patch area, vertical load and tire pressure to be able to calculate sinkage depth and relationate it with Crr. This relation considered that the pressure was constant over the contact patch but, as pressure peaks around the center of the contact patch and is zero in its limits, I was underestimating the area of the contact patch and, consequently, the sinkage depth for a given combination of normal load and tire pressure. This underestimation modified the relation between sinkage depth and Crr

In this new approach, I won't rely on contact patch surface to calculate sinkage depth. I will rather use wheel's vertical stiffness. The vertical stiffness of a tire is a product of air pressure principally because shoulder stiffness is very low. The relation between vertical stiffness and inflating pressure is sigmoidal but linear over typical values of pressure. Knowing that a road tire inflated up to 100psi has a vertical stiffness of 150000 N/m, we can calculate vertical stiffness for any pressure and determinate sinkage depth for a given load. After checking the results of the previous approach, I noticed that the sinkage depths were a bit too small. With this new method, the sinkage depths are 60-80% higher and they seem more plausible. This new approach gives the following relation between sinkage depth and Crr for the Specialized Turbo 25c clincher tire:

As you can see, the shape is still linear but the coefficients have changed

For this new relation, total power consumption in tire-ground contact for both rear ends showed in the previous post are the following ones:

Absolute value have changed but still 3.4W higher power consumption by increasing chainstay's diameter by just 8mm

Additionally, the flexibility of this new approach have allowed me to determinate the contact patch pressure function for an uncambered toroidal wheel. The shape of the pressure function proposed depends on the author of the paper that you read but it's typically considered to be unidimensional and sometimes doesn't consider the shift of the center of pressure caused by rolling resistance or it is said to change of shape depending on the slip. As Cossalter presented in one of his books and some FEA studies have endorsed, I'm inclined to think that the vertical pressure distribution is constant independently of the slip. Taking this into account, I propose the following bidimensional pressure distribution for a toroidal wheel:

R and r are the two defining radius of the torus. sk is the sinkage depth. a and b are two tuning parameters. x is the travelling direction. y is the perpendicular direction in the contact patch

This function assures that the pressure is null in the limits of the contact patch and allows the shift of the center of pressure to translate the rolling resistance. The tuning parameters a and b are negative and their modules increase when the sinkage depth is decreased.  The innovation of this approach is the obtention of these coefficients using experimental data of Crr (something similar could be done for the overturning moment). As the center of pressure is independent of the slip, the relation that says that Crr is the ratio between the shift of the center of pressure and the rolling radius for a wheel turning at constant angular speed is applicable. Knowing this, we can calculate the shift of the center of pressure for every inflating pressure (that modifies vertical stiffness and, consequently, sinkage depth) for a given load. The two equations that allow me to calculate both coefficients are 1) the integral of the pressure distribution over the contact patch is equal to the vertical load and 2) the center of pressure in the travelling direction is equal to the product of tire's external radius minus sinkage depth and Crr. This system of equations involves integrals over a complex area and can't be computed analitically. I have solved it for every pressure using a Mathematica+MAPLE combo (Mathematica to compute the integrals and MAPLE to solve the system numerically) because none of the programs could solve the whole problem. Some graphical representations of the pressure distribution:

R=0.3275m. r=0.0125m. sk=0.002556m. Pressure=127.91psi. Load=490.5N. Stiffness=191862.2N/m. Crr=0.004081

Contact patch for the previous conditions

Same conditions. Cut for y=0 to show the displacement of the center of pressure that translates the rolling resistance

R=0.3275m. r=0.0125m. sk=0.006544m. Pressure=49.97psi. Load=490.5N. Stiffness=74952.98N/m. Crr=0.007098

Contact patch for the previous conditions

That's all for today. Thanks for reading!

PD If you analyze the pressure distributions, you will see that the average pressure in the contact patch is approximately equal to the inflating pressure. There is an interesting relation between contact pressure, sinkage depth, Crr, inflating pressure and tire stiffness.

The return of the legend

After the latest regulation craze of the UCI, it's nice to hear these words:

Rim and disc brakes. Rear end vertical stiffness

In one of the many topics that there are currently in the forums about disc brakes on road bikes, I had a small discussion about the effects of the braking system in rear end's deformation. I was pretty sure that rim brakes generate less vertical deflection of the wheel axle during braking so I decided to write the equilibrium equations to confirm my thoughts.

To simplificate things, I have considered the rear end to be plane and formed by two identical tubes. I have also applied linearity (small strain) in the system to avoid having also to consider inertial and slipping forces due to deceleration. If something isn't clear, don't hesitate to ask

That's all for today. Thanks for reading!

PD If the disc brake force isn't parallel to the seatstays, the M moment is even smaller, so the vertical deflection would be bigger.

Wheel contact dynamics

Even though traction in the wheels is one the major performance improvements felt by the cyclist after changing his frame or wheel setup, it's not commonly highlighted by the marketing departments to try to sell a rigid frame. One of the reasons for this practice is that the abilty to maintain traction of a frame can't be as easily presented to the public as a BB stiffness test because there is a subtle line between a frame that gives good traction and one that is too flexy vertically. Altough this is the typical practice, some manufacturers highlight frame's ability to maintain traction. One recent example is the Cannondale SuperSix Evo presented last year. Cannondale has a long term experience using carbon properties to build softail frames with its ZeroPivot technology so it seems that they decided to apply some of this knowledge in the road side of things. Peter Denk developed a road version of this technology, called it Speed Save and said this about it:

 "We’re not using Speed Save for rider comfort, but more for the performance of the bike,” said Denk. “The suspension in a Formula One car isn't to keep the driver comfortable… [it's] to help keep the wheels and the tires connected to the track... We want the frame and fork to have some compliance to help the wheels stay in contact with the pavement better and to give that sensation of floating."

Let's analyze this paragraph in depth to understand if the loss of contact between the tire and the road can be detrimental to the performance of the bike. If the wheel loses contact with the ground, it will accelerate because there is no friction force and, consequently, the wheel will slip when it makes again contact with the ground because its linear velocity will be bigger than the linear velocity of the bicycle. Slipping means increased friction so we should consider both aspects of the loss of contact. If the energy disipated reducing wheel linear velocity to bicycle's velocity is bigger than the energy saved because the friction is 0 during the period of time that the wheel is not in contact with the ground, the loss of traction will be detrimental. We should also consider that if the cyclist can`t develop all his power because the bicycle is continually bouncing and he isn't comfortable, the reduced traction will be also detrimental. As you can imagine, evaluate this complex system takes some major engineering efforts, it's not an easy task.

I decided to start with something simple and add complexity later. The model that I've developed is a spring-damper system with one of the ends of the spring-damper following the road profile and the other one attached to a mass equivalent to the normal reaction force in the wheel that depends on the weight distribution of the bicycle. I have considered this equivalent mass to be constant but if I finally solve the dynamic equations presented in the Physics II post, I will be able to also consider the variations in the normal force induced by the pedalling motion. Another simplification that I've included for the moment is to consider the horizontal velocity of the bicycle constant. Considering that the road profile has a sinusoidal shape, the governing equation of the system is the following one:

Where N is the normal reaction force in flat terrain and static conditions, g is the acceleration of the gravity, c is the damping term, k the spring stiffness, R the radius of the unloaded wheel, A is the difference between the highest and lowest point of the road profile, w the angular frequency of the profile and v the horizontal velocity of the bicycle. x(t) and l(t) are the distance between the lowest point of the road profile and the wheel axle and the contact between the ground and the wheel respectively measured along the normal to the road profile in each point. As you can see there is no stiffness terms that depends on the frame/fork but I will try to include them after doing some future work about rear ends and forks stiffness.

Next step, do the first test. I have found some problems here to find stiffness and damping values of the wheel-tire sytem. Finally I have found a paper with these values of a road tire nevertheless I don't know neither what tire is nor the inflating pressure but I will use them as orientative values. Another thing to point out is that the model allows the spring to be longer that its unloaded length, something that it's impossible for a wheel. I will include later a variable stifness and damping that has a null value when the stretch is positive to solve this. To prevent the loss of contact between the wheel and the ground I have chosen the parameters cautiously. Finally, I have calculated the maximum angular frequency to ensure that there is only a contact patch between the tire and the ground in static conditions and I have used it for this first test, to do this I have considered that the distance between the center of the contact patch and the equivalent force of the pressure distribution in the contact is approximately equal to the semi-length of the contact patch. The values for this first test are the following ones:

And finally, the graphics. You can see that the maximum force is more than 1.6 times bigger than the static force. We'll see how frame's role affects this maximum value

The wheel and the road profile for the test, note the contact patch. The length of the irregularities is 9.2cm

That's all for today. Greetings

PD I have fixed a small error

Lateral loads

Hello all, I was short of time these last weeks so I couldn't update the blog. The next few weeks I will have some free time to try to develop some topics that I have in mind.

This post is dedicated to lateral loads in the contact between the wheel and the road, as many of you know, this type of load has a strong relation with the riding style of the cyclist because it is generated by the lateral motion of the bicycle and the rider. I have searched for bibliography about this subject and I have noticed that it's a topic that has been almost forgotten. I have only found some papers, the very good "Bicycle Wheel" and a very old but very technical one, both are mainly dedicated to the behaviour of the wheels when they are loaded laterally but there isn't a method to quantify these loads. You can also read Cervélo's engineer Damon Rinard's thoughts about this topic here and here. I have also noticed that the EN testing protocol doesn't include any fatigue test for lateral loads for the frames.

I decided to study a little bit the subject after following a topic created in Weight Weenies about the big test of wheels that Roues Artisanales does from time to time. In the 2008 test you can find a drawing of the forces acting in the wheel-ground contact and a small experimental explanation given by a Mavic's engineer. After working a bit with this diagram I noticed that the lateral loads in the plane of the wheel were zero so something must be wrong, this and the lack of an analytical model make me try to develop a proper one.

First, to understand better the mechanics of standing pedalling we can watch this small video where we can see Cancellara's start of a Tour of California TT:

After watching the video we can conclude that:

1) The lateral movement of the bicycle and the rider is approximatedly a zigzag so I will consider that there is no moment in the vertical direction

2) The lean angle of the bicycle is zero when the crank arms are parallel to the ground and maximum when one of the legs is in its lowest position

The last conclusion is not always true but I will use it to simplify things:

3) The cyclist is always in vertical position so his lean angle is always zero

This last conclusion is generally true for standing pedalling in flat terrrain but it depends a lot on the cyclist. For example, Cancellara accelerates during climbing with his body perfectly vertical while Voeckler moves a lot his body even accelerating in flat terrain. Taking this into consideration I developed the following model:

The following graphics show the results obtained for different values of the parameters and also the normal and lateral forces decomposed in the coincident and normal to the bicycle planes. I have to point out that the time origin is set when the crank arms are parallel to the ground. The results obtained seem to agree with the minimal level of lateral stiffness to avoid pad rubbing recommended by some experimental studies.

As I have already commented the distribution of this force between the two wheels depends on the variable weight distribution of the system. After solving the in-plane loads model that I presented some time ago, I will be able to calculate the three components of force loading each wheel and, consequently, the frame.

 That's all for today. Thanks for reading!

PD It's important to note that the lateral motion of the bicycle is more difficult to modelise, here I present a simplified model. It involves the lateral motion induced by the gravity during the first part of the movement and, when the lean angle approaches its maximum value, a progressive countersteering done by the cyclist to increase the height of the COG of the system and avoid falling. Due to the small sideslip angles involved, the difference between these forces and the real ones should be negligeable, specially for the rear wheel.