Riding a 26...

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  • 05-05-2004
    jimbo
    Riding a 26...
    I switched bikes with my ss g/f yesterday. I rode a pretty good ways on her Voodoo Nzumbi, while she rode my Mute. I actually really liked the way it handled. It felt quick to accelerate even with a bigger gear (in inches as well as ratio). It was quicker in the tight, twisties (which we have lots of here in Florida).

    I thought it was interesting. The Mute is certainly a juggernaut, but it is not a quick bike. I want to try both bikes on the same course a few times to get a feel for which one I am faster on. The Nzumbi certainly feels faster on the climbs, but I'm not sure if it is only a feeling, or reality.

    Here's one thought I had. Because 29'er wheels are both heavier and have the weight further from the center of rotation, the are slower to accelerate. When you are climbing you are constantly accelerating, even at a steady speed, because of gravity. So why would a 29'er climb faster than a 26" wheel bike? Maybe it's because the 29'er wheels to accelerate as much to maintain the same speed (fewer rpms).

    I'm not sure about all this, but I do know that the Nzumbi was FUN to ride. I'll be curious to here what other people have to say.

    jimbo
  • 05-06-2004
    Fattirewilly
    Quote:

    Originally Posted by jimbo
    I switched bikes with my ss g/f yesterday. I rode a pretty good ways on her Voodoo Nzumbi, while she rode my Mute. I actually really liked the way it handled. It felt quick to accelerate even with a bigger gear (in inches as well as ratio). It was quicker in the tight, twisties (which we have lots of here in Florida).

    I thought it was interesting. The Mute is certainly a juggernaut, but it is not a quick bike. I want to try both bikes on the same course a few times to get a feel for which one I am faster on. The Nzumbi certainly feels faster on the climbs, but I'm not sure if it is only a feeling, or reality.

    Here's one thought I had. Because 29'er wheels are both heavier and have the weight further from the center of rotation, the are slower to accelerate. When you are climbing you are constantly accelerating, even at a steady speed, because of gravity. So why would a 29'er climb faster than a 26" wheel bike? Maybe it's because the 29'er wheels to accelerate as much to maintain the same speed (fewer rpms).

    I'm not sure about all this, but I do know that the Nzumbi was FUN to ride. I'll be curious to here what other people have to say.

    jimbo


    What is this talk of riding a 26. Be gone with yee...:)

    Okay, now that it's out of my system, I'd have to say that any advantage a 29er has in climbing comes from the ability to roll over objects easier and the increased traction. That's it. If the climb is smooth, I think an equivalent 26 would be faster.

    I'm curious how tall you are? My wild guess is that you're under 6 foot and that contributed to your enjoyment of the 26 in the tight stuff.
  • 05-06-2004
    ncj01
    Quote:

    Originally Posted by Fattirewilly
    I'm curious how tall you are? My wild guess is that you're under 6 foot and that contributed to your enjoyment of the 26 in the tight stuff.

    At the risk of further offending Rhino and the growing list of folks angry over my photo posting, here is Jimbo, I'm guessing 5'11"-ish:

    <img src="https://picserver.org/view_image.php/ZC8V399Y1W9H/picserver.jpeg">
  • 05-06-2004
    appleSSeed
    ncj01, your pics haven't shown up for like 4 or 5 posts...haha, just thought I'd let you know
  • 05-06-2004
    Cloxxki
    The contant acc. during climbing is not true. Try lifting a fast turning wheel by the QRs, is it heavier than a still one? Answer : no!
  • 05-06-2004
    Fattirewilly
    Quote:

    Originally Posted by appleSSeed
    ncj01, your pics haven't shown up for like 4 or 5 posts...haha, just thought I'd let you know

    I can see them, with ms Int. Expl. At home with Mozilla, the pics aren't showning up.
  • 05-07-2004
    jimbo
    Not quite a workable analogy...
    Quote:

    Originally Posted by Cloxxki
    The contant acc. during climbing is not true. Try lifting a fast turning wheel by the QRs, is it heavier than a still one? Answer : no!

    No one has a perfectly smooth spin, sorry, no one. This is exaggerated when climbing. Because of this and the tendency of the bike to immediately slow down when climbing if any pressure at all is removed from the drivetrain, the bike is indeed in a constant acceleration mode.

    This is exacerbated by being on a singlespeed where you frequently have to stand while climbing. If you can't maintain a perfectly smooth spin on a gearie, you certainly can't do it standing up on an ss.

    This becomes even more true still on a technical climb where you are force to change speed and direction due to obstacles. Yes, a 29er makes it easier to go up and over the obstacles, but that difference is fairly small, and usually still involves acceleration.

    Everytime the wheel accelerates, the small wheels will accelerate faster.

    jimbo
  • 05-07-2004
    ncj01
    You're not accelerating when moving at a constant speed, up a hill or otherwise. Here's the link becuase I don't think the photo's are going to come through
    https://www.glenbrook.k12.il.us/gbss...Kin/U1L1e.html
    <html>
    <head>
    <meta http-equiv="Content-Type" content="text/html; charset=windows-1252">
    <meta name="GENERATOR" content="Microsoft FrontPage 4.0">
    <meta name="ProgId" content="FrontPage.Editor.Document">
    <title>New Page 1</title>
    </head>

    <body>

    <h2><font color="#ff0000">Lesson 1: Describing Motion with Words</font></h2>
    <h3><font color="#ff0000">Acceleration</font></h3>
    <p>The final mathematical quantity discussed in Lesson 1 is acceleration. An
    often confused quantity, acceleration has a meaning much different than the
    meaning associated with it by sports announcers and other individuals. The
    definition of acceleration is:<br>
    </p>
    <blockquote>
    <font color="#ff0000"><b>Acceleration</b></font> is a <a href="https://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L1b.html">vector
    quantity</a> which is defined as &quot;the rate at which an object changes its
    <a href="https://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L1d.html">velocity</a>.&quot;
    An object is accelerating if it is changing its velocity.
    <p>&nbsp;</p>
    </blockquote>
    <p><a name="vttable"></a><img src="https://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L1e1.gif" align="right" width="187" height="228"><a name="p2"></a>Sports
    announcers will occasionally say that a person is accelerating if he/she is
    moving fast. Yet acceleration has nothing to do with going fast. A person can be
    moving very fast, and still not be accelerating. Acceleration has to do with
    changing how fast an object is moving. If an object is not changing its
    velocity, then the object is not accelerating. The data at the right are
    representative of an accelerating object - the velocity is changing with respect
    to time. In fact, the velocity is changing by a constant amount - 10 m/s - in
    each second of time. Anytime an object's velocity is changing, that object is
    said to be accelerating; it has an acceleration.</p>
    <center><a href="https://www.glenbrook.k12.il.us/gbssci/phys/mmedia/kinema/acceln.html"><img src="https://www.glenbrook.k12.il.us/gbssci/phys/Class/animn.gif" align="bottom" border="0" width="95" height="25"></a>
    <p>&nbsp;</p>
    </center>
    <p>Sometimes an accelerating object will change its velocity by the same amount
    each second. As mentioned in the above paragraph, the data above show an object
    changing its velocity by 10 m/s in each consecutive second. This is referred to
    as a <font color="#ff0000"><b>constant acceleration</b></font> since the
    velocity is changing by a constant amount each second. An object with a constant
    acceleration should not be confused with an object with a constant velocity.
    Don't be fooled! If an object is changing its velocity -whether by a constant
    amount or a varying amount - then it is an accelerating object. And an object
    with a constant velocity is not accelerating. The data tables below depict
    motions of objects with a constant acceleration and a changing acceleration.
    Note that each object has a changing velocity.</p>
    <center><img src="https://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L1e12.gif" align="bottom" X-CLARIS-USEIMAGEHEIGHT X-CLARIS-USEIMAGEWIDTH width="367" height="205">
    <p>&nbsp;</p>
    </center>
    <p><a name="dtsquared"></a><a name="p3"></a>Since accelerating objects are
    constantly changing their velocity, one can say that the distance traveled/time
    is not a constant value. A falling object for instance usually accelerates as it
    falls. If we were to observe the motion of a <font color="#ff0000"><b>free-falling
    object</b></font> (<a href="https://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L5a.html">free
    fall motion</a> will be discussed in detail later), we would observe that the
    object averages a velocity of 5 m/s in the first second, 15 m/s in the second
    second, 25 m/s in the third second, 35 m/s in the fourth second, etc. Our
    free-falling object would be constantly accelerating. Given these average
    velocity values during each consecutive 1-second time interval, we could say
    that the object would fall 5 meters in the first second, 15 meters in the second
    second (for a total distance of 20 meters), 25 meters in the third second (for a
    total distance of 45 meters), 35 meters in the fourth second (for a total
    distance of 80 meters after four seconds). These numbers are summarized in the
    table below.</p>
    <center>&nbsp;
    <p>&nbsp;
    <table width="90%" border="3">
    <tbody>
    <tr>
    <td>
    <h3><font color="#ff0000"><b>Time Interval</b></font></h3>
    </td>
    <td>
    <h3><font color="#ff0000"><b>Ave. Velocity During Time Interval</b></font></h3>
    </td>
    <td>
    <h3><font color="#ff0000"><b>Distance Traveled During Time Interval</b></font></h3>
    </td>
    <td>
    <h3><font color="#ff0000"><b>Total Distance Traveled from 0 s to End of
    Time Interval</b></font></h3>
    </td>
    </tr>
    <tr>
    <td>0 - 1 s</td>
    <td>5 m/s</td>
    <td>5 m</td>
    <td>5 m</td>
    </tr>
    <tr>
    <td>1 -2 s</td>
    <td>15 m/s</td>
    <td>15 m</td>
    <td>20 m</td>
    </tr>
    <tr>
    <td>2 - 3</td>
    <td>25 m/s</td>
    <td>25 m</td>
    <td>45 m</td>
    </tr>
    <tr>
    <td>3 - 4 s</td>
    <td>35 m/s</td>
    <td>35 m</td>
    <td>80 m</td>
    </tr>
    </tbody>
    </table>
    </center>
    <p><a name="p4"></a>This discussion illustrates that a <a href="https://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L5a.html">free-falling
    object</a> which is accelerating at a constant rate will cover different
    distances in each consecutive second. Further analysis of the first and last
    columns of the data above reveal that there is a square relationship between the
    total distance traveled and the time of travel for an object starting from rest
    and moving with a constant acceleration. The total distance traveled is directly
    proportional to the square of the time. As such, if an object travels for twice
    the time, it will cover four times (2^2) the distance; the total distance
    traveled after two seconds is four times the total distance traveled after one
    second. If an object travels for three times the time, then it will cover nine
    times (3^2) the distance; the distance traveled after three seconds is nine
    times the distance traveled after one second. Finally, if an object travels for
    four times the time, then it will cover 16 times (4^2) the distance; the
    distance traveled after four seconds is 16 times the distance traveled after one
    second. For objects with a constant acceleration, the distance of travel is
    directly proportional to the square of the time of travel.</p>
    <p><a name="p5"></a>The acceleration of any object is calculated using the
    equation<br>
    </p>
    <center><img src="https://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L1e2.gif" align="bottom" width="263" height="38"></center>
    <p>This equation can be used to calculate the acceleration of the object whose
    motion is depicted by the <a href="#vttable">velocity-time data table</a> above.
    The velocity-time data in the table shows that the object has an acceleration of
    10 m/s/s. The calculation is shown below.<br>
    </p>
    <center><img src="https://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L1e3.gif" align="bottom" width="265" height="37"></center>
    <p><img src="https://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L1e4.gif" align="right" width="277" height="166"><a name="p6"></a>Acceleration
    values are expressed in units of velocity/time. Typical acceleration units
    include the following:</p>
    <p>&nbsp;
    <table width="90" border="0">
    <tbody>
    <tr>
    <td><center><font color="#ff0000"><b>m/s/s</b></font></center></td>
    </tr>
    <tr>
    <td><center><font color="#ff0000"><b>mi/hr/s</b></font></center></td>
    </tr>
    <tr>
    <td><center><font color="#ff0000"><b>km/hr/s</b></font></center></td>
    </tr>
    </tbody>
    </table>
    <center>&nbsp;</center>
    <p><a name="p7"></a>At first, these units are a little awkward to the newcomer
    to physics. Yet, they are very reasonable units when you begin to consider the
    definition and equation for acceleration. The reason for the units becomes
    obvious upon examination of the acceleration equation.</p>
    <center><img src="https://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L1e5.gif" align="bottom" width="95" height="38"></center>
    <p>Since acceleration is a velocity change over a time, the units on
    acceleration are velocity units divided by time units - thus (m/s)/s or
    (mi/hr)/s.</p>
    <p><a name="dirn"></a>Since acceleration is a <a href="https://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L1b.html">vector
    quantity</a>, it will always have a direction associated with it. The direction
    of the acceleration vector depends on two things:<br>
    </p>
    <ul>
    <li>whether the object is speeding up or slowing down
    <li>whether the object is moving in the + or - direction</li>
    </ul>
    <p>&nbsp;</p>
    <p><a name="Thumb"></a><a name="p8"></a>The general <font color="#ff0000">RULE
    OF THUMB</font> is:</p>
    <blockquote>
    If an object is slowing down, then its acceleration is in the opposite
    direction of its motion.
    </blockquote>
    <p>This <font color="#ff0000">RULE OF THUMB</font> can be applied to determine
    whether the sign of the acceleration of an object is positive or negative, right
    or left, up or down, etc. Consider the two data tables below. In each case, the
    acceleration of the object is in the &quot;+&quot; direction. In Example A, the
    object is moving in the positive direction (i.e., has a positive velocity) and
    is speeding up. When an object is speeding up, the acceleration is in the same
    direction as the velocity. Thus, this object has a <font color="#ff0000"><b>positive
    acceleration</b></font>. In Example B, the object is moving in the negative
    direction (i.e., has a negative velocity) and is slowing down. According to our <font color="#ff0000">RULE
    OF THUMB</font>, when an object is slowing down, the acceleration is in the
    opposite direction as the velocity. Thus, this object also has a <font color="#ff0000"><b>positive
    acceleration</b></font>.</p>
    <center><img src="https://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L1e6.gif" align="bottom" width="357" height="160"></center>
    <p>&nbsp;</p>
    <p><a name="p9"></a>This same <a href="#Thumb">RULE OF THUMB</a> can be applied
    to the motion of the objects represented in the two data data tables below. In
    each case, the acceleration of the object is in the &quot;-&quot; direction. In
    Example C, the object is moving in the positive direction (i.e., has a positive
    velocity) and is slowing down. According to our <font color="#ff0000">RULE OF
    THUMB</font>, when an object is slowing down, the acceleration is in the
    apposite direction as the velocity. Thus, this object has a <font color="#ff0000"><b>negative
    acceleration</b></font>. In Example D, the object is moving in the negative
    direction (i.e., has a negative velocity) and is speeding up. When an object is
    speeding up, the acceleration is in the same direction as the velocity. Thus,
    this object also has a <font color="#ff0000"><b>negative acceleration</b></font>.</p>
    <center><img src="https://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L1e7.gif" align="bottom" width="357" height="161">
    <p><a href="https://www.glenbrook.k12.il.us/gbssci/phys/mmedia/kinema/avd.html"><img src="https://www.glenbrook.k12.il.us/gbssci/phys/Class/animn.gif" align="bottom" border="0" width="95" height="25"></a></p>
    </center>
    <p><font color="#ff0000">&nbsp;</font></p>
    <p><font color="#ff0000">&nbsp;</font></p>
    <p><font color="#ff0000">&nbsp;</font></p>
    <h2><center><a name="cyu"></a><font color="#ff0000">Check Your Understanding</font></center></h2>
    <p>To test your understanding of the concept of acceleration, consider the
    following problems and the corresponding solutions. Use the equation for
    acceleration to determine the acceleration for the following two motions.</p>
    <center><img src="https://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L1e9.gif" align="bottom" width="368" height="141"></center>

    </body>

    </html>
  • 05-07-2004
    Fattirewilly
    Quote:

    Originally Posted by jimbo
    No one has a perfectly smooth spin, sorry, no one. This is exaggerated when climbing. Because of this and the tendency of the bike to immediately slow down when climbing if any pressure at all is removed from the drivetrain, the bike is indeed in a constant acceleration mode.

    This is exacerbated by being on a singlespeed where you frequently have to stand while climbing. If you can't maintain a perfectly smooth spin on a gearie, you certainly can't do it standing up on an ss.

    This becomes even more true still on a technical climb where you are force to change speed and direction due to obstacles. Yes, a 29er makes it easier to go up and over the obstacles, but that difference is fairly small, and usually still involves acceleration.

    Everytime the wheel accelerates, the small wheels will accelerate faster.

    jimbo

    Well the 29" wheel is harder to decelerate, so this would somewhat offset your theory of needing to accelerate with each turn of the crank, IMO.
  • 05-07-2004
    03bart
    It's not acceleration, it's force...
    True, if the velocity of an object is not changing, then there is no net accleration.

    However, that does not mean that there is no acceleration taking place.

    Your bicycle, riding on the flats, does not go forever with a single impluse of power. Make a single rotation of the pedals and your bicycle will go a short distance and stop.

    That's because it experienced a negative accleration due to forces acting on the bike. These forces include friction in the bearings, and friction and drag on the wheels. In order to maintain a constant velocity on the bike you have to continually apply a force in the forward direction by turning the pedals to overcome those forces acting against you.

    Now, start riding up a hill and you've added an additional force working against you, gravity.

    Force = mass x acceleration. Clearly, the more mass you're trying to move, the greater the force required to acheive the same acceleration. Now, rotating or angular forces work similarly but have an additional term proportional to the radius of the center of mass of the rotating object. So, the farther away the center of mass of the object from the center of rotation, the more force is required to achieve the same accleration.

    IOW, I don't know why the big wheels would feel better going up a hill than the small ones. But they do require more force from the rider, all else being equal.