# Calculating crank to diameter ratio ?'s

Hi,

When calculating the ratio of cranks to diameter, should I be trying to reach a ratio of 4.0?

For example: a crank ratio of 3.9037 would give me more torque than a ratio of 3.859?

Or do I have this concept reversed?

Can anyone detect this close a difference in torque in real world riding based on using the above numbers?

I calculated the diameter from the circumference of the tire using pi.

I then converted the diameter to mm and divided by crank length.
For example Using 160mm cranks and a diameter of 616.08mm
I divided 616.08 by 160 to get my ratio of 3.855.

Did I do this correctly?

If I did this correctly, a 24 x 2.6 Gazz with 160mm cranks is nearly the same as a 24 x 3.0 gazz with 170mm cranks. The difference is 4 thousandths. Gazz 24 x 3.0 is calculated as 656.08mm/170mm which equals 3.859.

Thanks

Re: Calculating crank to diameter ratio ?'s

On Sun, 15 Sep 2002 23:40:35 -0500, teachndad

>When calculating the ratio of cranks to diameter, should I be trying to
>reach a ratio of 4.0?
To begin with: why 4.0?
Secondly, you are talking about “the ratio of cranks to diameter” as
if it is a fixed notion, but is it? If not, I would rather use a
“crank to diameter” ratio.

>For example: a crank ratio of 3.9037 would give me more torque than a
>ratio of 3.859?
The way you express it, it would be the other way around. While you
call your number a “ratio of cranks to diameter”, what you actually
calculated is the ratio of diameter to cranks.

>Can anyone detect this close a difference in torque in real world riding
>based on using the above numbers?
It is barely 1%. I doubt if it would make a noticeable difference.
>
>I calculated the diameter from the circumference of the tire using pi.
>
>I then converted the diameter to mm and divided by crank length.
>For example Using 160mm cranks and a diameter of 616.08mm
>I divided 616.08 by 160 to get my ratio of 3.855.
>
>Did I do this correctly?
Yes, with the above proviso.

For your example, I would proceed as follows:
The radius of the wheel is 313.04 mm (half of the diameter). The
radius of the crank (a.k.a. crank length) is 160 mm. The crank to
diameter ratio then is 160/313.04, which equals 0.5111. That is sort
of a leverage factor, meaning that if you push with force x on the
pedal (perpendicular to the crank), then the force at the ground
contact point will be 0.5111 times x. So you would have more torque
(e.g. for climbing) if your crank to diameter ratio is higher
(assuming that you could still muster force x on the pedal).
Obviously, you can’t go higher than a ratio of 1, or your pedals would
protrude beyond the tyre. Bonk bonk bonk bonk. Indeed, they would bonk

Hope this helps,

Klaas Bil

If you had this signature, I have forged it.

Re: Calculating crank to diameter ratio ?'s

> Hope this helps,

Example:
I want to buy a 24" MUni and I want enough torque to make it up hills
with a minimal amount of effort. However, I don’t want the cranks so
long that my overall ride becomes jerky. I have asked myself, do I want
170 mm or 175 mm cranks? It seems to me that an important factor not
included in the equation is the size of the rider. If I were 6’-6" tall
with rather long legs then the ride would feel very different on 175 mm
cranks than with my shorter stature of 5’-10" tall. Until I actually
ride on the different cranks I won’t know anything other than the amount
of force represented by x, which doesn’t really help or does it?

Jason

It’s dead simple if you think of the cranks in inches - you think of the wheel in inches after all.
Roughly:
100 mm = 4 inches.
125mm = 5 inches
150 mm = 6 inches
170 mm is a shade under 7 inches.

Then it’s the radius of the wheel which matters, not the diameter, so simplify the calculation here too - the lever acts through the hub.

So, on a 20 inch wheel (10 inch radius) with 125 cranks (5 inches) you can express the leverage as 50% (If you prefer: 1:2)

On a 24 inch wheel (12 inch radius) with 6 inch cranks, the leverage is also 50%

This suggests the two set ups offer similar leverage - although there are many other factors which come into play.

On a Coker with a 36 inch wheel (18 inch radius) and standard 6 inch cranks, the ratio is 33% or 1:3.

Keep it simple… count the cranks in inches.

I found my Coker with 6 inch cranks would go up some hills my 20 wouldn’t go up with 5s on it. I found my 26 with 7 inch cranks (53%) would not go up stuff that my 24 with 6s (50%) would.

So, the ratio is a guide only, and the actual wheel size and actual crank size have an effect due to rolling resistance, momentum, angle of attack on obstacles, and ergonomics.

Re: Calculating crank to diameter ratio ?'s

I agree that other factors than just leverage play a role as per
Mikefule’s last sentence. He also simplifies the calculations by using
round inch numbers but the principle remains the same. And no, it’s
not academic. When I went from 24" with 125 mm cranks to 20" with 125
mm cranks, I noticed a definite improvement for uphilling. I mounted
150 mm cranks on my 24" wheel and hey presto, an improvement just as
big (compared to 125 mm). And guess what, the leverage factor is
(nominally) exactly the same.

Klaas Bil

If you had this signature, I have forged it.

On Mon, 16 Sep 2002 16:32:11 -0800, Jason Neumann
<nospam@nospam.no.no.no> wrote:

>
>Example:
>I want to buy a 24" MUni and I want enough torque to make it up hills
>with a minimal amount of effort. However, I don’t want the cranks so
>long that my overall ride becomes jerky. I have asked myself, do I want
>170 mm or 175 mm cranks? It seems to me that an important factor not
>included in the equation is the size of the rider. If I were 6’-6" tall
>with rather long legs then the ride would feel very different on 175 mm
>cranks than with my shorter stature of 5’-10" tall. Until I actually
>ride on the different cranks I won’t know anything other than the amount
>of force represented by x, which doesn’t really help or does it?
>
>
>Jason

Re: Calculating crank to diameter ratio ?'s

On Tue, 17 Sep 2002 14:01:14 -0500, Mikefule
<Mikefule.b51yy@timelimit.unicyclist.com> wrote:

>I found my Coker with 6 inch cranks would go up some hills my 20
>wouldn’t go up with 5s on it. I found my 26 with 7 inch cranks (53%)
>would not go up stuff that my 24 with 6s (50%) would.

For the second example, OK that’s a small difference in leverage
factor. Maybe also the 26" was a wider and hence larger tyre than the
nominal value? But the Coker example is remarkable, unless either
these were real short hills and you could use the greater speed
achievable with the Coker to carry you over the top, or the surface
was so bumpy that the small wheel got stuck.

Klaas Bil

If you had this signature, I have forged it.

Re: Calculating crank to diameter ratio ?'s

Klaas Bil wrote:
>
> I agree that other factors than just leverage play a role as per
> Mikefule’s last sentence.

Yes. After reading Mikefuls post I find this information much more
useful, expressing it in relative percentages such as, “50% (If you
prefer: 1:2)”.

Thanks guys!

Jason

There is still one more thing to here. Remember to measure the diameter of the tyre. For example, my 20" trainer has wheel diameter of 20" but my Gazza 24" has diameter of 26". This 24-26 diameter only affects 5%, but I think you may feel it.

I can’t seem to find it on his site right now, but quite a while back i downloaded a cranklength spreadsheet from Roger Davis…still use it a lot, it lets you plug in wheel diameter or circumference and crank length and come up with ratio and lots of other interesting stuff.

Chuck

Yep, the gaz 24x3 is about 26" (radius 13"), however, it’s typically run at such a low pressure, that I wonder if the effective radius (distance from center of hub to point of nearest contact) isn’t closer to 12.5" or even 12" - implying an effective 25" or 24" diameter.

.duaner.

actually, all that stuff is wrong. the center of rotation for rolling objects (wheels) is not the hub. it’s where the wheel touches the ground. so to you use the wheel radius plus the crank length instead of just the crank length. and no, i didn’t think of this on my own or make it up

Try www.unicycle.uk.com/Cranklength.xls I think

um, whoops. that shouldn’t be wheel radius + crank length. that should be distance of the pedal from the wheel’s contact point, which is wheel radius + crank length only when the crank is pointing up

I didn’t believe that, but thinking about it I think it’s true and it’s obvious that this is a different number and doesn’t average out to the same thing.

Why:-

When the pedals are at 9 o’clock, 12 o’clock and 3 o’clock, the distance between the point where force is being applied and the contact patch is greater than the wheel radius. It’s only in a small section of the pedal circle that the pedals are less than the wheel radius away from the ground and that doesn’t come near to balancing out the time spent in the other parts of the pedal circle.

Being currently a programmer and having forgotten my maths, I wrote a little program to work it out, for a 18 inch radius wheel (coker sized) with 18 inch cranks the effective radius is 22.9 inches, going down to 18.5 inches for 6 inch cranks and tending towards 18 inches as crank length goes to zero.

It doesn’t matter really though as it doesn’t make much difference to the calculations anyway, even in the most extreme practical cases (eg. 6 inch cranks on 20 inch wheel = effective radius of about 22 inches).

I’m not sure what happens on a bike, although the small size of rear sprockets might mean it’s only a very tiny amount of constant difference in gearing due to the drive being from slightly behind the axle.

Joe

Re: Calculating crank to diameter ratio ?'s

On Sat, 27 Sep 2003 22:00:43 -0500, ubersquish
<ubersquish.ug44z@timelimit.unicyclist.com> wrote:

>actually, all that stuff is wrong. the center of rotation for rolling
>objects (wheels) is not the hub. it’s where the wheel touches the
>ground. so to you use the wheel radius plus the crank length instead of
>just the crank length.
>[if the pedal is at 12 o’clock]

That only holds in a reference frame fixed to the road. So it only
applies to someone standing (or walking) on the road. Indeed, if you
stand beside a unicycle and want to propel it forward, pushing against
the top pedal gives you most leverage, equal to wheel radius plus
crank length.

However, if you think about a unicyclist riding a unicycle, then you
have to consider rotation in a reference frame fixed to the unicycle.
Then the centre of rotation is at the hub, and the usual notions about
leverage apply.

## Klaas Bil - Newsgroup Addict

If the crank is moving then it really sounds as if it’s loose. - onewheeldave trying to pinpoint the cause of a clicking crank

Thanks Klaas Bil. I’m impressed. Take that Joe Marshall. No offense, I just had to say that for no reason. I’m sorry. I just kept thinking and thinking over what you said about counting cranks and wheel radius together.

nope, the center of rotation is still where the wheel meets the road. who said it had to always be the same spot? it moves. and that’s what integrals are for. ooh, the calculus

Yippee!! I guess all the people in the world just love, integrals. They are so nice.

Integrals and calculus are wonderful, however, they are totally unnecessary for the problem at hand. The center of rotation of interest in this basic problem is the hub, not the point of ground contact.

.duaner.

Okay, so only of interest to people with a particular obsession with pushing their unicycle then.

What I thought first was that the pedals are effectively a lever with a fulcrum at the centre of the wheel and a constant length due to the direction of the force going in being perpendicular to the cranks. So I’m not really sure how I thought the rolling around the contact patch thing would apply.

Joe