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OT: What simple rule of math am I missing here?

In the grand scheme of things, this certainly isn't important. But it's something that has been bugging me, mainly because I think there's something obvious I'm missing.

When I run on my treadmill and set the speed at 5 mph, it sets a pace of a 12-minute mile, so I run ... 5 miles in an hour.

But why is it, when I alternate quarter-miles at 4 mph, then 6 mph, then 4, then 6, etc., it takes 62-plus minutes to cover 5 miles? Shouldn't 10 quarter-miles at 4 mph and 10 at 6 mph average out to 5 mph?

(See disclaimer at top about the lack of importance.)

Rob

Columbus, Ohio

«13

Comments

  • HeavyG
    HeavyG Posts: 10,323
    If Bob is on a train heading west at 125mph and Sally is on a train heading east at 150mph...
    “Reality is that which, when you stop believing in it, doesn't go away.” ― Philip K. Diçk




  • nolaegghead
    nolaegghead Posts: 42,102
    Seems about right.
    ______________________________________________
    I love lamp..
  • 4 mph = 15 min mile
    6 mph = 10 min mile

    (15+10)/2 = 12.5 * 5 miles = 62.5 min

    youre mixing up your units. 
  • maso
    maso Posts: 240
    10 quarters is 2.5 miles. 

    2.5 miles at 4mph pace takes 37.5 min. 
    2.5 miles at 6mph pace takes 25 min. 

    Add together. 62.5 min. 
    Large BGE in Moore, OK
  • YukonRon
    YukonRon Posts: 16,984
    Are these preset quarters, or do you manually change these paces during your workout? The time to go to 4mph to 6mph and reverse may add the additional time to complete the process.
    Best guess.
    "Knowledge is Good" - Emil Faber

    XL and MM
    Louisville, Kentucky
  • 4 mph = 15 min mile
    6 mph = 10 min mile

    (15+10)/2 = 12.5 * 5 miles = 62.5 min

    youre mixing up your units. 
    Thanks! Seeing it like you laid it out makes perfect sense, so I do get it.

    What I can't get my ahead around is why two equal distances, one covered at 4 mph and the other at 6 mph, doesn't average at 5.

    This explains why I bailed on math after Algebra II and went into journalism.

    Rob

    Columbus, Ohio

  • maso
    maso Posts: 240
    You can only average rates like that if they're executed for the same amount of time. 

    So 4mph for 1 hour is 4 miles. 
    Then 6mph for an hour is 6 miles

    total of 10 miles over 2 hours averages 5 mph. 
    Large BGE in Moore, OK
  • DoubleEgger
    DoubleEgger Posts: 17,125
    4 mph = 15 min mile
    6 mph = 10 min mile

    (15+10)/2 = 12.5 * 5 miles = 62.5 min

    youre mixing up your units. 
    Good ol' GT education 
  • maso said:
    You can only average rates like that if they're executed for the same amount of time. 

    So 4mph for 1 hour is 4 miles. 
    Then 6mph for an hour is 6 miles

    total of 10 miles over 2 hours averages 5 mph. 
    Because the units (HR) cancel each other out. Gotta solve for the units. 
  • 42?
    Simi Valley, California
    LBGE, PBC, Annova, SMOBot
  • Wolfie51sb
    Wolfie51sb Posts: 267
    edited September 2017
    maso said:
    You can only average rates like that if they're executed for the same amount of time. 

    So 4mph for 1 hour is 4 miles. 
    Then 6mph for an hour is 6 miles

    total of 10 miles over 2 hours averages 5 mph. 
    Boom! That made it click. Thanks.

    Thanks, also, @theyolksonyou.

    Rob

    Columbus, Ohio

  • thetrim
    thetrim Posts: 11,352
    Let's wait for the math professor to chime in.  
    =======================================
    XL 6/06, Mini 6/12, L 10/12, Mini #2 12/14 MiniMax 3/16 Large #2 11/20 Legacy from my FIL - RIP
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    EIB 6 Oct 95
  • 1voyager
    1voyager Posts: 1,157
    thetrim said:
    Let's wait for the math professor to chime in.  
    I'm too busy smoking Pueblo chiles.
    Large Egg, PGS A40 gasser.
  • JohnInCarolina
    JohnInCarolina Posts: 30,864
    edited September 2017
    Velocity is distance over time. So time is distance over velocity.

    The original time is one hour.  If you want that to remain the same using two different velocities that just happen to average to your original 5mph, then you will have to walk *different* distances for each of them.  Let's see if we can show this.

    The original velocity is 5 miles per hour.  So we have:

    1 = (5mi)/(5 mph) 

    or

    T = D/V 

    We want the same T for two different Ds and two different Vs, or

    T = D1/V1 + D2/V2, where D1+D2 = D.  

    In your case, the V1 (6mph) is 1.2 V, and the V2 (4mph) is 0.8V.  So we have

    T = D1/ 1.2V + D2/ 0.8V

    or

    T = D1/1.2V + 1.5 D2/ 1.2V = (D1 + 1.5D2) / 1.2V

    So this will be the same as the above if

    (D1 + 1.5D2)/1.2 = D

    and we also know that:

    (D1 + D2) = D.  

    I'm assuming you all know how to solve a simultaneous system of linear equations for two unknowns, so I'll leave it there.   But suffice it to say that it is easy to show that D1 = D/2 and D2 = D/2 is not a solution.  


    "I've made a note never to piss you two off." - Stike
  • :rofl:  dude hates math and you post this shît
  • I will say that the previous answers were also fine, they just didn't provide a general solution.  
    "I've made a note never to piss you two off." - Stike
  • :rofl:  dude hates math and you post this shît
    He was happy with the answer he had. This was for the rest of you clowns. =)
    "I've made a note never to piss you two off." - Stike
  • bgebrent
    bgebrent Posts: 19,636
    Velocity is distance over time. So time is distance over velocity.

    The original time is one hour.  If you want that to remain the same using two different velocities that just happen to average to your original 5mph, then you will have to walk *different* distances for each of them.  Let's see if we can show this.

    The original velocity is 5 miles per hour.  So we have:

    1 = (5mi)/(5 mph) 

    or

    T = D/V 

    We want the same T for two different Ds and two different Vs, or

    T = D1/V1 + D2/V2, where D1+D2 = D.  

    In your case, the V1 (6mph) is 1.2 V, and the V2 (4mph) is 0.8V.  So we have

    T = D1/ 1.2V + D2/ 0.8V

    or

    T = D1/1.2V + 1.5 D2/ 1.2V = (D1 + 1.5D2) / 1.2V

    So this will be the same as the above if

    (D1 + 1.5D2)/1.2 = D

    and we also know that:

    (D1 + D2) = D.  

    I'm assuming you all know how to solve a simultaneous system of linear equations for two unknowns, so I'll leave it there.   But suffice it to say that it is easy to show that D1 = D/2 and D2 = D/2 is not a solution.  


    That's the most pedantic solution imaginable. While I know it's right, John for the win. Or is your name "Mathboy"?
    Sandy Springs & Dawsonville Ga
  • @bgebrent - you can just call me "Your Excellency"  
    "I've made a note never to piss you two off." - Stike
  • DoubleEgger
    DoubleEgger Posts: 17,125
    @bgebrent - you can just call me "Your Excellency"  
    I prefer Mathalete 
  • @bgebrent - you can just call me "Your Excellency"  
    I prefer Mathalete 
    Not "Mathole" ?
    "I've made a note never to piss you two off." - Stike
  • bgebrent
    bgebrent Posts: 19,636
    I like Mathboy for the dookie. 
    Sandy Springs & Dawsonville Ga
  • @bgebrent - you can just call me "Your Excellency"  
    I prefer Mathalete 
    I will (somewhat) proudly say that somewhere in my mom's house, I have some math competition trophies. Ok, after seeing that, not that proud. 
  • JohnInCarolina
    JohnInCarolina Posts: 30,864
    edited September 2017
    @bgebrent - you can just call me "Your Excellency"  
    I prefer Mathalete 
    I will (somewhat) proudly say that somewhere in my mom's house, I have some math competition trophies. Ok, after seeing that, not that proud. 
    Yeah. They stopped letting me compete in those at some point because all of the other parents kept complaining.  I started signing in as i raised to the ith power to throw them off, but it didn't work.  
    "I've made a note never to piss you two off." - Stike
  • bgebrent
    bgebrent Posts: 19,636
    I started out as a math major. It became clear to me very early on that a chemistry major would get hotter chicks. It's all about plastics. And math majors sucked at physical chemistry. 
    Sandy Springs & Dawsonville Ga
  • SmokeyPitt
    SmokeyPitt Posts: 10,490
    I think you got it already...but since you are running a fixed distance segments (1/4 mile), you are spending more time running at the 4 MPH rate so it brings the average rate down. If you were to instead use time segments, e.g. run 5 minutes at 4 mph, 5 minutes at 6 mph then the average would be 5 MPH. 


    Which came first the chicken or the egg?  I egged the chicken and then I ate his leg. 

  • maso
    maso Posts: 240
    Velocity is distance over time. So time is distance over velocity.

    The original time is one hour.  If you want that to remain the same using two different velocities that just happen to average to your original 5mph, then you will have to walk *different* distances for each of them.  Let's see if we can show this.

    The original velocity is 5 miles per hour.  So we have:

    1 = (5mi)/(5 mph) 

    or

    T = D/V 

    We want the same T for two different Ds and two different Vs, or

    T = D1/V1 + D2/V2, where D1+D2 = D.  

    In your case, the V1 (6mph) is 1.2 V, and the V2 (4mph) is 0.8V.  So we have

    T = D1/ 1.2V + D2/ 0.8V

    or

    T = D1/1.2V + 1.5 D2/ 1.2V = (D1 + 1.5D2) / 1.2V

    So this will be the same as the above if

    (D1 + 1.5D2)/1.2 = D

    and we also know that:

    (D1 + D2) = D.  

    I'm assuming you all know how to solve a simultaneous system of linear equations for two unknowns, so I'll leave it there.   But suffice it to say that it is easy to show that D1 = D/2 and D2 = D/2 is not a solution.  


    I mean that's wrong but ok. 
    Large BGE in Moore, OK
  • Lit
    Lit Posts: 9,053
    What's a treadmill?
  • bgebrent said:
    I started out as a math major. It became clear to me very early on that a chemistry major would get hotter chicks. It's all about plastics. And math majors sucked at physical chemistry. 
    Worked out for you. 
  • maso
    maso Posts: 240
    maso said:
    Velocity is distance over time. So time is distance over velocity.

    The original time is one hour.  If you want that to remain the same using two different velocities that just happen to average to your original 5mph, then you will have to walk *different* distances for each of them.  Let's see if we can show this.

    The original velocity is 5 miles per hour.  So we have:

    1 = (5mi)/(5 mph) 

    or

    T = D/V 

    We want the same T for two different Ds and two different Vs, or

    T = D1/V1 + D2/V2, where D1+D2 = D.  

    In your case, the V1 (6mph) is 1.2 V, and the V2 (4mph) is 0.8V.  So we have

    T = D1/ 1.2V + D2/ 0.8V

    or

    T = D1/1.2V + 1.5 D2/ 1.2V = (D1 + 1.5D2) / 1.2V

    So this will be the same as the above if

    (D1 + 1.5D2)/1.2 = D

    and we also know that:

    (D1 + D2) = D.  

    I'm assuming you all know how to solve a simultaneous system of linear equations for two unknowns, so I'll leave it there.   But suffice it to say that it is easy to show that D1 = D/2 and D2 = D/2 is not a solution.  


    I mean that's wrong but ok. 
    IF you want an formula for average speed, with each speed over a same distance, then:

    Speed = 2ab/(a + b), where a and b are the speeds. so in this situation, a = 4, b = 6 so

    2*4*6/(4+6) = 48/10 or 4.8 mph (or 5 miles in 62.5min)
    Large BGE in Moore, OK