MooseFET said:
The user said that this was the "range".
No, he did not. He said he did 12 miles at 17 mph
comfortably. He did not say 12 miles is the range.
And why do you think that range = total energy in
the battery? I've got a couple of 12V lead acid
car batteries that won't crank over the car starter,
but happily run a headlamp for over an hour at ~6.5
amps. We don't know th state of charge of his
battery so it is impossible to know that all of
the enery was extracted.
Any energy not going to make
the range is a loss of some form or other.
Completely ambiguous, and it is not a closed
system. For example, energy expended to fight a
headwind is energy going to make the range, so
it doesn't fit into "energy not going to make
the range". It is not energy expended in a
non-productive way, but it is a drain on the
total energy. So it is ambiguous as to whether
you call that a loss or not, but the net effect
is that more energy is required.
If you are doing so, you are incurring a loss.
If you are doing what?
This is what the OP said. Are you suggesting he lied?
No. He said:
"At the moment, it does 12 miles at 17 mph quite comfortably
with no input from the rider."
He did not say "the range was *limited* to 12 miles."
You assumed "limited".
The OP said that it was the range and spoke of a charger.
He did *not* say it was the range.
And what does the part about he "spoke of a charger"
have to do with it? Go read the post. He said:
"Just curious to what you think of this idea of extending the range.
A leisure battery from a scrap yard 12v 110 amp - £30
A fast charger 22amp from Argos - £40
A 12v to 36 dc to dc convertor - £70
A current limiting diode."
No you just have taken too narrow of a view of what is called a loss.
That may be true. But it does not change my
objection to the assumptions that the battery
was fully capable of meeting its 12AH rating;
that the discharge rate matched the 12 AH spec;
that the full 12 AH was extracted from the battery.
Now if all of that is lumped under "loss", then the
statement that there are huge losses in the system
is true. But it is meaningless under that wide of
a meaning of "loss". For example, suppose the 12AH
battery is old and weak, and the best it can do is
6 AH. "Your battery may be weak" is a whole lot more
helpful than "there are huge losses in the system".
Aside from that, the glaring error is performing
the calculations against 12AH when you don't know
the rate of discharge and the battery's capacity
at that rate. You do know, don't you, that batteries
discharged at a rate above their specified AH will
provide lower total AH at the higher rate?
Once again "paleface" comment pops to mind.
So provide figures. Your comment is just a smokescreen.
Define very little. Have you ever measured? or even estimated it?
You must have never ridden a bike to take exception to the
statement. I've ridden it, and measured speed. But measuring
is just another smokescreen. The difference in speed and effort
is so big that no metrology is required to know it.
But set that aside - look at world class cyclists in a race.
When they are climbing mountains, they go at a snail's
pace as compared to ridng on level roads and have to expend
a lot more energy and develop a lot more power climbing.
Watch the Tour du France - the commentators will mention
speed and effort. In the Gossamer Albatross, hp was measured,
and considerably more was required in turbulent air vs still air.
Ed
