Ganapathy and I attempted climbing Basavanabetta using single speed ratio of 43:16. On the way, we two were calculating the maximum gradient one can do using a given gear ratio or the amount of force required to climb a given gradient. I tried googling for any such existing article or formula. Since I couldn't find one, I'm writing it down here if anybody wants to refer.

I have deliberately excluded the possibility of using power in the upward stroke as I hate cleats!

#### What is gradient?

It is height gained divided by distance covered horizontally multiplied by 100.

gradient = (height gained / horizontal distance) * 100

#### What does it take to climb or push an object?

Many articles seem to be confused between power and force (linear) or torque (rotational). Power helps in moving faster. Torque helps in rolling against a resistance (gravity or wind or whatever). Of course, power and torque are related. But the problem comes when people try to use a power meter and calculate their max power on a flat road and then try to deduce how much gradient they would be able to do. What is the problem? On flat the rider would have reached the max power in a gear ratio that allows him to spin a an optimum cadence, say, 100 rpm. That cannot be applied when we want to calculate the maxi gradient one can climb up in a given gear ratio, because at that limit, the cadence tends to zero. It will be zero when he just manages to prevent a rolling backward. That is the limit we are interested to calculate. Now, you see why power is useless here.

Another simple example if you remember physics is that when you lift a heavy bag and walk on the flat road, you do zero work because gravitational force is downward and displacement is horizontal. No work means zero power! But we constantly apply a force upward on the bag so that it does not fall to the ground.

#### Final gear ratio

A rider applies the force on the pedal. We will assume that he always applies the pressure tangential to the crank arm to avoid loss of the force. This force is then transferred onto the wheel at the contact point with the road, which makes the bike eventually move. The final gear ratio,

*r*, from tire to pedal is:*r = (rear wheel radius /*

*crank arm length*

*) **

*(*

*number of chain ring teeth /*

*number of rear cog teeth*

*)*

#### What is the maximum gradient a given gear ratio can climb?

Lets assume the gradient is

*x*. We will find out the angle of the road to the horizontal level, say,*θ*. Then,

*θ =*

*tan*

^{-}^{1}*x*

Now, lets find out the force that drags the bike backward on the up slope. This force will be applied tangentially at the contact point of the tires. Lets assume the weight of rider and bike as

*w*. The force that drags backward,*d*, will be equal to:

*d = w * sin*

*θ*

Here, we are going to assume infinite amount of grip to avoid tire skidding downward. Now, the reverse force on the pedal due to the up slope, say,

*p*, which is tangential to the crank arm, can be defined as:*p = r * d*

*=>*

*p = sin(*

*tan*

^{-}^{1 }*gradient) * (rider weight + bike weight) **

*(rear wheel radius / crank arm length) **

*(*

*number of chain ring teeth /*

*number of rear cog teeth*

*)*

Lets take the example of my attempt on Basavanabetta with 43:16 ratio with 170 mm crank arm and 700x25c tires. Average gradient of Basavanabetta is 10%, but I will take 15% to accommodate the steeper sections, which will be the challenging part.

The reverse force on the pedal will be:

*p = sin (*

*tan*

^{-}^{1}*(0.15)) * ((59 kg + 15 kg) * g) **

*(335 mm / 170 mm) * (43 teeth / 16 teeth)*

*= 58 kg * g*

*where, g is the acceleration due to gravity.*

I didn't evaluate the acceleration due to gravity to keep the force in humanly conceivable number. Another reason I didn't evaluate is that I'm going to find out if I will be able to prevent the bike from rolling back with full body weight on the pedal when the crank arm is horizontal to the ground. At that moment, the force on the pedal is also my weight multiplied by

*g*. Since my weight is 59 kg and*p*is 58 kg, I will just be able to hold onto the bike without rolling it back. But I will not be able to pedal the bike up the slope. Even if I lower the gradient significantly, I will be able to pedal up only at 3 o'clock position of the crank arm and will be forced to roll backward as I won't be able to put my full weight equivalent force perpendicular to the crank arm.
Oh, I didn't give the formula for the max gradient. That is easy now. If we assume the max gradient as the limit at which the bike just stops rolling up when the rider is standing on the pedal and crank arm is horizontal to the ground, then

*m = p*

*where, m is the rider weight*

*=> m = r * d*

*= r * w * sin*

*θ*

*sin*

*θ = m / r / w*

*Hence,*

*θ =*

*sin*

^{-}^{1}*(m / r / w)*

*=*

*sin*

^{-}^{1}*{m * (crank arm length / rear wheel radius) **

*(number of rear cog teeth / number of chain ring teeth) /*

*w}*

*Then,*

*gradient = tan*

*θ * 100*

If we assume a road bike with smaller ring as 28 and largest cog as 28, rider as 70 kg and bike weight as 12 kg, then the max gradient angle,

*θ = sin*

^{-}^{1}*{70 * (170 / 335) * (28 / 28) / (70 + 12)}*

*=*

*sin*

^{-}^{1}*(0.433)*

*= 25.67 degree*

*Hence,*

*gradient = tan (25.67 degree) * 100*

*= 48%*

Please remember the caveat I gave :) This is the gradient at which the rider will just be able to hold the bike from rolling back with its full body weight on the pedal and crank arm being horizontal to be able to leverage the gravity.

One can deduce the minimum gear ratio to climb a given gradient and weight.

One can deduce the minimum gear ratio to climb a given gradient and weight.

In summary, what helps in climbing higher gradient? Longer crank arm, smaller chain ring, larger rear cog, smaller tire diameter and a lighter bike of course :)

I have deliberately excluded the possibility of using power in the upward stroke as I hate cleats!