Cycling Equipment · Public discussion

Math question (Trainer)

Started by Hell and High Water · · Last activity · 26 posts · 1,077 views

This thread is locked and is currently read-only.

Thread navigation

Jump through the discussion

Go to the original post, the replies on this page, or the latest preserved contribution.

Thread details

What we know about this thread

Original section
Cycling Equipment
Published
11 October 2005
Last activity
12 October 2005
Original author
Hell and High Water
Posts
26
Discussion status
Public discussion
Total views
1,077
Views / 30 days
0

The navigation and discussion metadata provide context. Posts remain in their original chronological order.

Showing posts 1–20 of 26
Posts remain in their original chronological order.

Text size
  1. I'm on my trainer and I'm pedaling exactly 100 RPM.

    Front chainring is 53

    Rear cog is 14

    Tires are 700 x 23

    Trainer is set to the middle of three resistance settings

    If I was doing this on the road, how fast would I be travelling?

    (seriously....this isn't a trick question. I just don't know how to
    figure it out...)

    THANKS!

    -Bob

  2. Hell and High Water said:

    I'm on my trainer and I'm pedaling exactly 100 RPM.

    Front chainring is 53

    Rear cog is 14

    Tires are 700 x 23

    Trainer is set to the middle of three resistance settings

    If I was doing this on the road, how fast would I be travelling?

    Quick and dirty approximate answer:
    (Chainring size/cog size)*(pedal cadence (in rpm))*.08
    gives speed in mph.

    For your numbers, 53/14*100*.08 ~= 30mph.

    Mark

  3. "Hell and High Water" wrote: If I was doing this on the road, how fast would
    I be travelling? (clip)
    ^^^^^^^^^^^^^^^
    The resistance setting doesn't matter. I get pretty close to 30 MPH.
    However, called the wheel diameter 27"--don't know the correct diameter of a
    700 wheel.

    Multiply pedal RPM by chain ring teeth, and divide by cog teeth. That gives
    you wheel RPM. Then calculate circumference of wheel (pi x diameter) and
    multiply to get distance per minute. If diameter was given in inches, then
    you will get speed in inches per minute. So divide by 12 and by 5280 to get
    miles per minute. x 60 gives miles/hour.

  4. Hell and High Water said:

    I'm on my trainer and I'm pedaling exactly 100 RPM.
    Front chainring is 53
    Rear cog is 14
    Tires are 700 x 23
    If I was doing this on the road, how fast would I be travelling?
    THANKS!

    -Bob

    You would be going 26.66 mph

    Using this calculator recalculate for MPH at Bottom after fist
    calculation> http://www.panix.com/~jbarrm/cycal/cycal.30f.html

    Based on your wheel size being 668mm in diameter. Based on ISO formula
    at bottom of Sheldon's page.
    http://sheldonbrown.com/cyclecomputer-calibration.html

    John

  5. Hell and High Water said:

    I'm on my trainer and I'm pedaling exactly 100 RPM.
    Front chainring is 53
    Rear cog is 14
    Tires are 700 x 23 > Trainer is set to the middle of three resistance settings
    If I was doing this on the road, how fast would I be travelling?

    Quoted message said:

    -Bob

    You would be going 29.62MPH at 100 rpms

    See here> http://www.panix.com/~jbarrm/cycal/cycal.30f.html

  6. Hell and High Water said:

    I'm on my trainer and I'm pedaling exactly 100 RPM.

    Front chainring is 53

    Rear cog is 14

    Tires are 700 x 23

    Trainer is set to the middle of three resistance settings

    If I was doing this on the road, how fast would I be travelling?

    Without some objective gauge of the resistance ofthe trainer, it's
    impossible to say. If you used a powermeter on your bike while on the
    trainer to get a reading, then went outside and rode the same rpm and
    same power output (likely with a different gear) you could approximate
    the speed.

    JT

    ****************************
    Remove "remove" to reply
    Visit http://www.jt10000.com
    ****************************

  7. "Hell and High Water" <[email hidden]> wrote in message
    news:[email hidden]...

    Quoted message said:

    I'm on my trainer and I'm pedaling exactly 100 RPM.

    Front chainring is 53

    Rear cog is 14

    Tires are 700 x 23

    Trainer is set to the middle of three resistance settings

    If I was doing this on the road, how fast would I be travelling?

    (seriously....this isn't a trick question. I just don't know how to
    figure it out...)

    (82.5/12)*(53/14)*(100/60)*(60/88) = 29.57 mph.

    Rolling circumference of 70x23 tire = 82.5 inches
    53/14 = gear ratio
    100/60 = rev/sec
    60/88 converts ft/sec to mph.

    Phil H

  8. Hell and High Water said:

    I'm on my trainer and I'm pedaling exactly 100 RPM.

    Front chainring is 53

    Rear cog is 14

    Tires are 700 x 23

    Trainer is set to the middle of three resistance settings

    If I was doing this on the road, how fast would I be travelling?

    It's impossible to make even an educated guess with the information.
    Most of the answers so far, I believe, missed the point. Your rear tyre
    is spinning at 29.6 mph, but to estimate how fast you would be going on
    the road with the same level of EFFORT we need to know what your
    POWER output is.

    Assuming that you're trainer doesn't have a display showing wattage,
    it's difficult to tell. Have you checked if your trainer's manual has a
    graph showing the wattage given speed with the three resistance
    settings?

    -as

  9. In article <[email hidden]>,
    [email hidden] says...

    Quoted message said:

    It's impossible to make even an educated guess with the information.
    Most of the answers so far, I believe, missed the point. Your rear tyre
    is spinning at 29.6 mph, but to estimate how fast you would be going on
    the road with the same level of EFFORT we need to know what your
    POWER output is.

    Effort has nothing to do with it.

    If I'm in a measurable gear, pedaling at a measurable rate, I'll be
    travelling at a measurable speed.

    It's mathematics.

    Now...If I'm out on the road, there's a lot more to being able to pedal
    at that cadence in that gear, but if I DID IT, I'd be travelling at a
    certain speed.

    -Bob

  10. In article <[email hidden]>, piholmanz@yourservice
    says...

    Quoted message said:

    (82.5/12)*(53/14)*(100/60)*(60/88) = 29.57 mph.

    Rolling circumference of 70x23 tire = 82.5 inches
    53/14 = gear ratio
    100/60 = rev/sec
    60/88 converts ft/sec to mph.

    This one looks right to me.

    700 x 23 is 82.5 inches?

    If that's correct, this makes sense.

    THANKS!

    -Bob

  11. Hell and High Water said:

    If I'm in a measurable gear, pedaling at a measurable rate, I'll be
    travelling at a measurable speed.

    It's mathematics.

    Nice troll.

    JT

    ****************************
    Remove "remove" to reply
    Visit http://www.jt10000.com
    ****************************

  12. Hell and High Water said:

    I'm on my trainer and I'm pedaling exactly 100 RPM.

    Front chainring is 53

    Rear cog is 14

    Tires are 700 x 23

    Trainer is set to the middle of three resistance settings

    If I was doing this on the road, how fast would I be travelling?

    (seriously....this isn't a trick question. I just don't know how to
    figure it out...)

    THANKS!

    -Bob

    Some great answers have been given here, but it depends on which
    is the dominant factor - your cadence or the power you're putting
    out. If the cadence dominates, the power will fluctuate with
    grade and wind resistance, etc, and most of the math here is
    correct. If the power is kept constant, the cadence and speed
    will be a function of grade and wind.

    p.

  13. In article <[email hidden]>,
    [email hidden] says...

    Quoted message said:
    Hell and High Water said:

    If I'm in a measurable gear, pedaling at a measurable rate, I'll be
    travelling at a measurable speed.

    It's mathematics.

    Nice troll.

    ??

    Not sure I get that one...

    -Bob

  14. Hell and High Water said:

    I'm on my trainer and I'm pedaling exactly 100 RPM.

    Front chainring is 53

    Rear cog is 14

    Tires are 700 x 23

    Trainer is set to the middle of three resistance settings

    If I was doing this on the road, how fast would I be travelling?

    Well, if you mean how fast you'd be going with that rpm:

    (53/14 * 100 Rotations Per Minute * 60 Minutes/hour *
    ((622 + 2 * 23 millimeters rim diameter)* pi) /10^6= kilometers per hour.
    Google for (53/14 * 100 * 60 * ((622 + 2 * 23) * pi )/10^6 (removing the
    units google calc doesn't understand) and it tells you the answer is
    47.67.

    On the other hand, the middle of three resistance settings almost
    certainly provides much less resistance than rolling resistance and wind
    resistance would out on the road, so your actual power output, measured
    against the actual road, would probably be quite a bit slower.

    Jasper

  15. In article <[email hidden]>, [email hidden]
    says...

    Quoted message said:

    Some great answers have been given here, but it depends on which
    is the dominant factor - your cadence or the power you're putting
    out. If the cadence dominates, the power will fluctuate with
    grade and wind resistance, etc, and most of the math here is
    correct.

    Exactly.

    I was specific in that I was pedaling at exactly 100 rpm.

    If I was in the exact same gear, pedaling the exact same rpm, I'd be
    traveling at a given, measurable speed. Regardless of how hard or soft
    I was pushing on the pedals.

    The gearing is set. One pedal revolution equates to (x)tire revolution.
    There would not be any 'slippage.'

    (82.5/12)*(53/14)*(100/60)*(60/88) = 29.57 mph.

    Rolling circumference of 70x23 tire = 82.5 inches
    53/14 = gear ratio
    100/60 = rev/sec
    60/88 converts ft/sec to mph.

    (I believe that answer is correct.)

    -Bob

  16. Hell and High Water said:

    Effort has nothing to do with it.

    If I'm in a measurable gear, pedaling at a measurable rate, I'll be
    travelling at a measurable speed.

    It's mathematics.

    Now...If I'm out on the road, there's a lot more to being able to pedal
    at that cadence in that gear, but if I DID IT, I'd be travelling at a
    certain speed.

    -Bob

    There are a number of programs that can calculate the info you
    supplied and do much more. One I really like is GearCalc, here take a
    look: http://www.income-software.com/index.htm?GearCalc/gc_index.asp

    It is real helpful for planning a drivetrain, it includes gear ratios,
    gear inches, cadence info, and much more.

    Life is Good!
    Jeff

  17. Jasper Janssen said:

    tells you the answer is 47.67.

    That's 47.67 km/hour. Or about 29.8 mph.

    The easy to remember formula is:

    Gear Ratio * Wheel Diameter (inches) * Cadence (rpm) * 0.00297 = MPH

    53/14 * 26.5 inches * 100 * 0.00297 = 29.8 mph

    The 0.00297 is the fudge factor that brings all the units into line.

    Art Harris

  18. In article <[email hidden]>,
    [email hidden] says...

    Quoted message said:
    Jasper Janssen said:

    tells you the answer is 47.67.

    That's 47.67 km/hour. Or about 29.8 mph.

    The easy to remember formula is:

    Gear Ratio * Wheel Diameter (inches) * Cadence (rpm) * 0.00297 = MPH

    53/14 * 26.5 inches * 100 * 0.00297 = 29.8 mph

    The 0.00297 is the fudge factor that brings all the units into line.

    On a side note, I went out for a ride this morning, dropped it into that
    exact gear, 53x14, pedaled with the same tune in my head, Stones 'Beast
    of Burden', looked down at the speedo and it was showing 29.

    SCIENCE!!

    ;-)

    -Bob

  19. If I was doing this on the road, how fast would I be travelling?
    ^^^^^^^^^^^^^^^
    It's impossible to make even an educated guess with the information. (clip)
    ^^^^^^^^^^^^^^^
    I can do better than an educated guess--I can give you an EXACT ANSWER. You
    would not be moving at all. The trainer stays put, no matter how fast you
    pedal or where it is located. ;-)

  20. "Hell and High Water" wrote: (clip) looked down at the speedo and it was
    showing 29.

    Quoted message said:


    SCIENCE!!


    ^^^^^^^^^^^^^^^^^^^^
    I think you should enter those results in a science fair. Can you help me
    design an experiment that verifies that 2 + 2 = 4? <G>

Active in the last 60 minutes

Active in this thread

0 users · 0 guests ·0 bots ·0 total

No signed-in users are active right now.

No known search crawlers active right now.