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

Wolfie51sb

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.)

HeavyG

If Bob is on a train heading west at 125mph and Sally is on a train heading east at 150mph...

nolaegghead

Seems about right.

theyolksonyou

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

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. 

YukonRon

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.

Wolfie51sb

theyolksonyou said:

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.

maso

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. 

DoubleEgger

theyolksonyou said:

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 

theyolksonyou

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. 

SoCal_Griller

42?

Wolfie51sb

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.

thetrim

Let's wait for the math professor to chime in.  

1voyager

thetrim said:

Let's wait for the math professor to chime in.  

I'm too busy smoking Pueblo chiles.

JohnInCarolina

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.  

theyolksonyou

 dude hates math and you post this shît

JohnInCarolina

I will say that the previous answers were also fine, they just didn't provide a general solution.  

JohnInCarolina

theyolksonyou said:

 dude hates math and you post this shît

He was happy with the answer he had. This was for the rest of you clowns.

bgebrent

JohnInCarolina 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.  

That's the most pedantic solution imaginable. While I know it's right, John for the win. Or is your name "Mathboy"?

JohnInCarolina

@bgebrent - you can just call me "Your Excellency"  

DoubleEgger

JohnInCarolina said:

@bgebrent - you can just call me "Your Excellency"  

I prefer Mathalete 

JohnInCarolina

DoubleEgger said:

JohnInCarolina said:

@bgebrent - you can just call me "Your Excellency"  

I prefer Mathalete 

Not "Mathole" ?

bgebrent

I like Mathboy for the dookie. 

theyolksonyou

DoubleEgger said:

JohnInCarolina said:

@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

theyolksonyou said:

DoubleEgger said:

JohnInCarolina said:

@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.  

bgebrent

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. 

SmokeyPitt

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. 

maso

JohnInCarolina 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. 

Lit

What's a treadmill?

theyolksonyou

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 said:

JohnInCarolina 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)