Big Green Egg - EGGhead Forum - The Ultimate Cooking Experience...
Welcome to the EGGhead Forum - a great place to visit and packed with tips and EGGspert advice! You can also join the conversation and get more information and amazing kamado recipes by following Big Green Egg at:

Facebook  |  Twitter  |  Instagram  |  Pinterest  |  Youtube  |  Vimeo
Share your photos by tagging us and using the hashtag #EGGhead4Life.


In Atlanta? Come visit Big Green Egg headquarters, including our retail showroom, the History of the EGG Museum and Culinary Center!  3786 DeKalb Technology Parkway, Atlanta, GA 30340.

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

  • HeavyGHeavyG Posts: 5,036
    If Bob is on a train heading west at 125mph and Sally is on a train heading east at 150mph...
    Camped out in the (757/804)
  • nolaeggheadnolaegghead Posts: 27,204
    Seems about right.
    ______________________________________________
    This is my signature line just so you're not confused.  Love me or hate me, I am forum Marmite.
    Large and Medium BGE, Kamado Joe Jr, Akorn Jr, smoker with a 5k btu AC, gas grill, fire pit, pack of angry cats, two turntables and a microphone, my friend.  Registered republican.
    New Orleans, LA - we know how to eat 

  • 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. 
    Jason NW GA- home of carpet and Mexican restaurants
    LBGE, MM, BS (Blackstone and the other kind)
    One sorry Labrador

    My chili did not suck. 
  • masomaso 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
  • YukonRonYukonRon Posts: 12,423
    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

  • masomaso 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
  • DoubleEggerDoubleEgger Posts: 12,170
    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. 
    Jason NW GA- home of carpet and Mexican restaurants
    LBGE, MM, BS (Blackstone and the other kind)
    One sorry Labrador

    My chili did not suck. 
  • 42?
    Simi Valley, California
    LBGE, PBC, Annova, SMOBot
  • Wolfie51sbWolfie51sb Posts: 261
    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

  • thetrimthetrim Posts: 6,090
    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
    Tampa Bay, FL
    EIB 6 Oct 95
  • thetrim said:
    Let's wait for the math professor to chime in.  
    I'm too busy smoking Pueblo chiles.
    Somewhere in Colorado
    LBGE, PGS A40 Gasser
  • JohnInCarolinaJohnInCarolina Posts: 11,831
    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.  


    "If the world is something you accept rather than interpret, then you're susceptible to the influence of charismatic idiots." - NdGT

    "The truth is, these are not very bright guys, and things got out of hand." - DT


  • :rofl:  dude hates math and you post this shît
    Jason NW GA- home of carpet and Mexican restaurants
    LBGE, MM, BS (Blackstone and the other kind)
    One sorry Labrador

    My chili did not suck. 
  • I will say that the previous answers were also fine, they just didn't provide a general solution.  
    "If the world is something you accept rather than interpret, then you're susceptible to the influence of charismatic idiots." - NdGT

    "The truth is, these are not very bright guys, and things got out of hand." - DT


  • :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. =)
    "If the world is something you accept rather than interpret, then you're susceptible to the influence of charismatic idiots." - NdGT

    "The truth is, these are not very bright guys, and things got out of hand." - DT


  • bgebrentbgebrent Posts: 16,625
    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"  
    "If the world is something you accept rather than interpret, then you're susceptible to the influence of charismatic idiots." - NdGT

    "The truth is, these are not very bright guys, and things got out of hand." - DT


  • DoubleEggerDoubleEgger Posts: 12,170
    @bgebrent - you can just call me "Your Excellency"  
    I prefer Mathalete 
  • @bgebrent - you can just call me "Your Excellency"  
    I prefer Mathalete 
    Not "Mathole" ?
    "If the world is something you accept rather than interpret, then you're susceptible to the influence of charismatic idiots." - NdGT

    "The truth is, these are not very bright guys, and things got out of hand." - DT


  • bgebrentbgebrent Posts: 16,625
    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. 
    Jason NW GA- home of carpet and Mexican restaurants
    LBGE, MM, BS (Blackstone and the other kind)
    One sorry Labrador

    My chili did not suck. 
  • JohnInCarolinaJohnInCarolina Posts: 11,831
    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.  
    "If the world is something you accept rather than interpret, then you're susceptible to the influence of charismatic idiots." - NdGT

    "The truth is, these are not very bright guys, and things got out of hand." - DT


  • bgebrentbgebrent Posts: 16,625
    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
  • SmokeyPittSmokeyPitt Posts: 9,885
    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. 

  • masomaso 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
  • LitLit Posts: 6,947
    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. 
    Jason NW GA- home of carpet and Mexican restaurants
    LBGE, MM, BS (Blackstone and the other kind)
    One sorry Labrador

    My chili did not suck. 
  • masomaso 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
Sign In or Register to comment.
Click here for Forum Use Guidelines.