maths problem !

[berty bassett]

Could anybody tell me the way to work out this maths problem . I go out for a 40 mile ride - first 10 mile is with someone pottering along chatting with av14.8 then i branch off and do the rest and when i get home the av is 18.4 - whats the formula to work out the average on last 30 . please help as i am sure its an easy sum as soon as i know how to do it , but i cant think of a way other than going in strava and cropping the ride and i dont really want to do that - thanks

 

[stargazer]

Total distance / total average = total time
First section distance / first section average = first section time

Total distance – first section distance = second section distance
Total time – first section time = second section time

Second section distance / second section time = second section average

 

[Cycleops]

Buy a cycle computer.

 

[Iain M Norman]

Split the ride into 4 averages. You know the 1st quarter's average and you know the resulting average for all four.

That final average is the average of all four so is all four added together and divided by four.

Assume that the last 3 are all the same, and they are the average unknown speed you want to know. Call it y.

So:

(14.8 + 3y) / 4 = 18.4

You should be able to solve y there with basic O/GCSE level maths. I'm not gonna do all the work for you. And you did ask for the formula.

*cheeky encouraging grin*

 

[e-rider]

19.6 mph

 

[HLaB]

Ive really spread out the itterations but:


Distance 10 & 30 = 40miles
Proportion of ride 0.25 & 0.75 = 1
Speed 14.8 & x = 18.4mph

Itteration
1) (.25*14.8) + 0.75x = 18.4
2) 0.75x = 18.4 - (.25*14.8)
3) 0.75x = 14.7
4) x = 14.7/.75
5) x = 19.6 mph

 

[Rasmus]

You've actually done slightly better than predicted by HLaB and Iain. You cannot weight averages with a distance.

In general terms,

d1/a1+d2/a2=(d1+d2)/atot

which after a bit of fiddling comes out as

a2 = atot / (d1/d2(1-atot/a1)+1)

plug in numbers and you get 20.02 mph

 

[Asa Post]

Rasmus is spot on, but I prefer to explain in English rather than Algebraic
I've rounded the figures to 2 decimal places, which causes a small difference between my result and his.

First 10 miles is done at an average speed of 14.8 mph. Time equals distance divided by speed, so the time for this is 0.68 hours.

Total ride is 40 miles at an average speed of 18.4 mph. By the same method as above, work out the total time, and it comes to 2.17 hours.

Now deduct the first 10 miles from the distance total, and the time for the first 10 miles from the time total. Distance equals 40 minus 10, giving 30 miles (duh). Time equals 2.17 minus 0.68, giving 1.49 hours.

So you have covered the last 30 miles in 1.49 hours. Speed equals distance divided by time, so the speed for this section of the ride equals 20.13 mph.

 

[Eribiste]

.....And >20 mph ain't so dusty!

 

[byegad]

Why care?

 

[MickeyBlueEyes]

Why care?

Because data is King......

 

[Leaway2]

http://www.mapmyride.com/ If you take a smart phone.

 

[byegad]

Because data is King......

Data?
What about enjoying the ride for the ride's sake?

 

[Rob3rt]

 

[Shut Up Legs]

My advice is: stick with the highest of the above figures, and reject all others as embarrassingly low .