CycloComputer Question

Hell and High Water said:

Is the average calculated by a factor of distance or time?

Both. Speed is distance/time (Miles/Hour).

The computer counts pulses from the wheel sensor. The computer also has
a clock. So "x" pulses in "y" seconds corresponds to some speed in mph
or kph. With most computers, the clock stops when there are no pulses
being received (i.e., the bike is stopped).

So, average speed is total distance divided by the total time the bike
was in motion.

Art Harris

Hell and High Water said:

Is the average calculated by a factor of distance or time?

IE:

(Theory here)

If I go up a one mile hill at 10 MPH, then come down the hill at 20 MPH,
my average will be 15 MPH. (Easy math)

OR....

If I go up a hill at 10 MPH for three minutes, then come down the hill
at 20 MPH for 30 seconds, my average is x. (Too lazy to do the math)

Does the computer average the amount of the DISTANCES I covered, or the
two TIMES I covered.

Again, theory here. No need to bring in the 'time to turn
around/slowing/speeding up/etc.' stuff.

TIA,

-Bob

speed = distance/time

"Francesco Devittori" wrote: speed = distance/time
^^^^^^^^^^^^^^^^^^
The OP asked a legitimate question, which I will rephrase here: Is the
calculated average a time based average or a distance based average. The
answer is: time based. The alternative, distance based average can be
calculated, but it is not useful.

Leo Lichtman said:

"Francesco Devittori" wrote: speed = distance/time
^^^^^^^^^^^^^^^^^^
The OP asked a legitimate question, which I will rephrase here: Is the
calculated average a time based average or a distance based average. The
answer is: time based. The alternative, distance based average can be
calculated, but it is not useful.

Sorry, I can't see the difference between my answer and yours, except form.

Hell and High Water said:

Is the average calculated by a factor of distance or time?

IE:

(Theory here)

If I go up a one mile hill at 10 MPH, then come down the hill at 20 MPH,
my average will be 15 MPH. (Easy math)

OR....

If I go up a hill at 10 MPH for three minutes, then come down the hill
at 20 MPH for 30 seconds, my average is x. (Too lazy to do the math)

Does the computer average the amount of the DISTANCES I covered, or the
two TIMES I covered.

Again, theory here. No need to bring in the 'time to turn
around/slowing/speeding up/etc.' stuff.

TIA,

-Bob

I think no "average" of seperate values is calculated.

The Average Speed = (Total distance) / (total time).

It is neither "time" nor "distance" based - it just is what it is.

Is your agerage earnings based on "time" or "money" ? ? ?

--
1) Eat Till SATISFIED, Not STUFFED... Atkins repeated 9 times in the book
2) Exercise: It's Non-Negotiable..... Chapter 22 title, Atkins book
3) Don't Diet Without Supplimental Nutrients... Chapter 23 title, Atkins
book
4) A sensible eating plan, and follow it. (Atkins, Self Made or Other)

"jbuch" wrote: (clip) It is neither "time" nor "distance" based - it just
is what it is.(clip)
^^^^^^^^^^^^^^^^^
I think you are missing the very distinction the OP was asking about. At
any moment, the bike is traveling with some instantaneous speed. If you
want to calculate an average speed, you have to sum up all the speeds along
the route, and divide by something--this can be either a total time, or a
total distance.

Suppose you ride up a one-mile hill at 4 mph, and then ride down at 30 mph.
What is your average speed for the round trip. A time-based average would
say: 15 minutes going up hill; 2 min going down hill, for a total time of
17 min, so average speed = 2 mi/17 min or .118 miles/min, or7 MPH.

A distance based average would say you traveled equal distances at two
speeds, so you just take their average: 4 + 30 = 34. Divide this by two
for an average of 17 MPH.

The second one is of hardly any interest, but let's try to cobble up a
possible way it might be useful. Let's say you wanted to sum up the total
energy consumed in all your bearings over the trip. And let's say the
bearing friction is proportional to speed. If you calculated a DISTANCE
average speed, you would have a number that might be useful in correlating
bearing energy losses.

"Francesco Devittori" wrote: Sorry, I can't see the difference between my
answer and yours, except form.
^^^^^^^^^^^^^^^
Because there is ANOTHER average that can be calculated, which is averaged
over distance, and the OP was asking which average the cyclometer displays.
I'm going to try to answer this more fully in my response to jbush.

In article <[email hidden]>,
[email hidden] says...

Quoted message said:

"jbuch" wrote: (clip) It is neither "time" nor "distance" based - it just
is what it is.(clip)
^^^^^^^^^^^^^^^^^
I think you are missing the very distinction the OP was asking about. At
any moment, the bike is traveling with some instantaneous speed. If you
want to calculate an average speed, you have to sum up all the speeds along
the route, and divide by something--this can be either a total time, or a
total distance.

Suppose you ride up a one-mile hill at 4 mph, and then ride down at 30 mph.
What is your average speed for the round trip. A time-based average would
say: 15 minutes going up hill; 2 min going down hill, for a total time of
17 min, so average speed = 2 mi/17 min or .118 miles/min, or7 MPH.

A distance based average would say you traveled equal distances at two
speeds, so you just take their average: 4 + 30 = 34. Divide this by two
for an average of 17 MPH.

See, that's exactly what I was wondering.

I went up the hill at 10, then down at 20. The hill was the same
distance, up and down, so my average should be 15. (or so I thought)

It doesn't work that way though. My average is WAY lower.

It seems like my average is always being lowered, simply because when
I'm moving slower, either into the wind, up a hill, whatever, it LASTS
so much longer.

When I'm flying down a hill at 30 MPH, it's over in 10 seconds and my
average speed was virtually unaffected....

-Bob (Patently stating the obvious since 1960)

Leo Lichtman said:

I think you are missing the very distinction the OP was asking about.
At any moment, the bike is traveling with some instantaneous speed.
If you want to calculate an average speed, you have to sum up all the
speeds along the route, and divide by something--this can be either a
total time, or a total distance.

If you use the definition, average speed = total distance divided
by the total time, then answer has to be "time based" or more precisely,
"time weighted".

Here's one for you:

A cyclist rides from Point A to Point B at 10mph. How fast does he have
to ride from Point B back to Point A (same route) to average 20mph for
the entire trip?

Art Harris

In article <[email hidden]>,
[email hidden] says...

Quoted message said:

Here's one for you:

A cyclist rides from Point A to Point B at 10mph. How fast does he have
to ride from Point B back to Point A (same route) to average 20mph for
the entire trip?

My first thought is 30mph, but that's probably not right...

-Bob

Hell and High Water said:

In article <[email hidden]>,
[email hidden] says...

Quoted message said:

Here's one for you:

A cyclist rides from Point A to Point B at 10mph. How fast does he have
to ride from Point B back to Point A (same route) to average 20mph for
the entire trip?

My first thought is 30mph, but that's probably not right...

Scotty, beam me over!

Infinite speed is needed...

For instance, say it is 10 miles in each direction... so it takes 1 hr
for the first leg... but to average 20mph you need to do the whole 20
miles in 1 hr!

That's why a hilly course will always hurt your average speed.

Ron Ruff said:

Scotty, beam me over! Infinite speed is needed...

Correct!

Art Harris

Hell and High Water said:

Is the average calculated by a factor of distance or time?

Neither: it's total distance traveled divided by total time taken.

Jasper

Art Harris said:

Here's one for you:

A cyclist rides from Point A to Point B at 10mph. How fast does he have
to ride from Point B back to Point A (same route) to average 20mph for
the entire trip?

The speed of light.

Jasper

Hell and High Water said:

In article <[email hidden]>,
[email hidden] says...

Quoted message said:

"jbuch" wrote: (clip) It is neither "time" nor "distance" based - it just
is what it is.(clip)
^^^^^^^^^^^^^^^^^
I think you are missing the very distinction the OP was asking about. At
any moment, the bike is traveling with some instantaneous speed. If you
want to calculate an average speed, you have to sum up all the speeds along
the route, and divide by something--this can be either a total time, or a
total distance.

Suppose you ride up a one-mile hill at 4 mph, and then ride down at 30 mph.
What is your average speed for the round trip. A time-based average would
say: 15 minutes going up hill; 2 min going down hill, for a total time of
17 min, so average speed = 2 mi/17 min or .118 miles/min, or7 MPH.

A distance based average would say you traveled equal distances at two
speeds, so you just take their average: 4 + 30 = 34. Divide this by two
for an average of 17 MPH.

See, that's exactly what I was wondering.

I went up the hill at 10, then down at 20. The hill was the same
distance, up and down, so my average should be 15. (or so I thought)

It doesn't work that way though. My average is WAY lower.

Quoted message said:

It seems like my average is always being lowered, simply because when
I'm moving slower, either into the wind, up a hill, whatever, it LASTS
so much longer.

When I'm flying down a hill at 30 MPH, it's over in 10 seconds and my
average speed was virtually unaffected....

Average Velocity = Distance you travel / time it takes [period]

For your hill:
time = 1 mile uphill / 10 mph + 1 mile downhill / 20 mph
time = 0.1 hour + 0.05 hours = 0.15 hours total

Ave. Vel. = (1 mile + 1 mile)/0.15 hours = 13.3 mph

--
Paul M. Hobson
Georgia Institute of Technology
..:change the words to numbers
if you want to reply to me:.

Jasper Janssen said:

On 11 Oct 2005 12:01:50 -0700, "Art Harris" <[email hidden]> wrote:

Quoted message said:

Quoted message said:

A cyclist rides from Point A to Point B at 10mph. How fast does he have
to ride from Point B back to Point A (same route) to average 20mph for
the entire trip?

The speed of light.

Nope, even several MegaLights won't do it. Instantaneous transfer from B
to A is the only way.

Mike

"Paul Hobson" wrote: (clip) Average Velocity = Distance you travel / time
it takes [period] (clip)
^^^^^^^^^^^^^^
No, Paul, not [period]. I'm not going to explain it again, but if you would
re-read some of the other posts in this thread, maybe you would realize that
the time-weighted average which you specify is not the only one. The OP
compared two DIFFERENT averages, asking which one the cyclometer computes.
The other one could also have some validity under other circumstances.
{period}.

Mike Causer said:

Jasper Janssen said:

On 11 Oct 2005 12:01:50 -0700, "Art Harris" <[email hidden]> wrote:

Quoted message said:

Quoted message said:

A cyclist rides from Point A to Point B at 10mph. How fast does he have
to ride from Point B back to Point A (same route) to average 20mph for
the entire trip?

The speed of light.

Nope, even several MegaLights won't do it. Instantaneous transfer from B
to A is the only way.

Ah, but due to the power of relativity, c essentially *is* instantaneous.

Jasper