Sport Rider Shift Point Atricle - Questions

Hi all,

This article posted quite a while ago is good at explaination on when to shift.
http://www.sportrider.com/tech/146_0402_art/index.html

I have one question though that i'm hoping someone will have some insight to;

Basically the part where you are calculating the driving force -
Force = Torque x Gear Ratio x Primary Reduction Ratio x Final Drive Ratio x Radius of Rear Wheel (include tyre)

Now I get all that except for the radius of the rear wheel. I know that the size of your wheel effects the torque multiplication of your engine (basically what your gear box does).

But when you open the excel file to see how they have used that calculation it doesn't make any sense to me.

This is what i think it does;
1 - calculates the radius off the circumference (easy)
2 - Then plugs that into a 'magic' formula that has parts not explained.

Shift Rpm = "Cross-over" Point (you select from the graph) / ((2 x Pi x Radius) / (12 x (60/5280)) x Gear Ratio x Primary Reduction Ratio x Final Drive Ratio)

To simply Shift Rpm = C/O Point / ((Circumference / 12 x (60/5820)) * Total Gear ratios)

where did the 12 x (60/5820) come from???? I'm assuming the 60 has got to do with revolutions per minute??

My estimate is that (Circumference / 12 x (60/5820) converts a wheel size to a multiplication ratio for the torque out of the engine, but what is it a ratio in comparison of (gear ratios are in the form 1st gear - 3.125:1 but we only write them as 1st gear - 3.125)

any help at all would be greatly appreciated as i'm building a car version of the same thing. The only difference being it estimates your torque curve based on peak kW & pear torque (not entirely accurate i know, but if you havn't got a dyno graph it's as close as you can get mathematically). My version will have the option to edit the estimated torque curve values if you have a dyno graph of your engine however.

cheers
ike  

anyone guys...

almost 70 views and no replies....

does anyone know how i can get in contact with Don Smith (author)???  

just re-read the article and excel file....

I miss read the formula. It should be;

Shift Rpm = "Cross-over" Point (you select from the graph) / ((2 x Pi x Radius) / 12) x (60/5280) x Gear Ratio x Primary Reduction Ratio x Final Drive Ratio)

Simplfied to;
Shift Rpm = C/O Point / (Circumference(feet) x (60/5820) x Total Gear ratios)


but that still leaves me with where did 60/5820 come from???



elsewhere in the excel file he uses 10/5820 as part of a formula to work out mph for each rpm.

btw. I'm from australia so it may be just a simple imperial conversion factor which i'm unaware of (we use metric in aus)  

Hey, haven't read the article in question but there are 5280 feet in a mile, and 60 minutes in an hour. Therefore it seems to me 60/5280 is just a conversion from feet per minute to miles per hour. Hope that helps, -Max  

thanks maxgeissler!!!   that sounds like it could be it....

ignore the typo's on my behalf (5280 NOT 5820)


i'm still not sure on the logic however....
if I have a dyno graph of my car typically 4th gear(approx 1:1 ratio) with a diff ratio of 4.35 and a wheel circumference of ~5feet (small wheels i know).

At 4500rpm it's putting 50ft/lbs of torque to the road (massive i know)

road force = torque x gear ratio x diff ratio x wheel

road force = 50 x 1 x 4.35 x wheel

what i want to do is find out what 'wheel' is (more accurately how wheel is calculated) so that i can find out what the road force is for different sized wheels.  

just looked at the road force formula and it seems backwards....


If i had a dyno graph (wheel dyno not engine dyno) of my vehicle wouldn't 50ft/lbs be the actual road force?? something like this...

road force = torque x gear ratio x diff ratio x wheel
50 = torque x 1 x 4.35 x wheel

as the dyno is measuring what is actually coming out of the wheels which is the same as the road force. Or am i missing something with dyno graphs where they convert it back to engine torque for the customer print out????