Heated gear- less MPGs?

[Knifemaker]

I guess I need someone to explain this to me. On another forum, a member posted that running heated gear lowers your miles per gallon.

I assumed that the stator is always putting out power, so there is no extra "drain" on the engine....thus fuel mileage would be not effected.

How would running more electrical accessories increase the load on the motor to effect fuel consumption?

 

[odot]

More then you know but less then an air conditioning unit.

 

[bgross]

"But officer, I HAD to turn my lights off - I was running low on fuel."

 

[ahchiu]

This is a first, and it's Friday.

 

[RedPorscheKilla]

How would running more electrical accessories increase the load on the motor to effect fuel consumption?

Hey Knifemaker,

long time no speak nor write. I hope all is good with ya.

Let me try to explain that one, coming from a cages life.....Our FeeJay's electric system is build to suffice the supply of electricity during "normal" 'n then some.... Now depending on that "and then some" will add to the normal consumption. Meaning the stator has a plus tolerance to supply enough electricity, usually from my experience about 15 - 25% from the initial max load for a short period of time. Usually you don't want to put more on a stator the the max output, then you'll drain that sucker to death at one point, which the usual sign is, that the battery won't get charged properly. (remember them driving low 'n slow with a gazillion watts of amplifiers in the trunk doing 24/7 the boom boom bang? They actually supply their cage with either an XL version of a stator or in an absolute ueber case a second stator, just to get by to supply enough electricity for that boom boom bang...).

So now back to your heated gear....the higher the amperage asked for the more the system has to work. Kind of compared to a usual motor having to cope with an a/c, power steering and what not. The more you add to the initial load, the more the motor will need to work in order to compensate for that, meaning since all is getting used more, because the added amps will be taken away from the stator, so the motor will sort of yell, I need more amps stator you sucker!! this is a cycle, in which you will eventually experience a higher usage of fuel subsequently lesser mileage per gallon. This shouldn't be more than 5 - 10% lesser fuel mileage though.

Hope I made sense with my layman's talk

V

RPK

 

[jcyuhn]

A permanent magnet alternator (i.e. magneto) such as the FJR uses always runs at full output. The regulator/rectifier, for lack of a better word, dumps the excess power produced to keep the electrical system operating at the correct voltage. Use more power and the RR dumps less, use less power and the RR disposes of more. Hence I would not expect using heated clothing to have any direct affect on the fuel economy of the FJR. I would expect to get lower fuel economy riding in colder temperatures where it is nice to have heated gear.

An alternator that uses induction to generate the magnetic field (i.e. any car alternator) will consume more fuel to generate more power. There are a few motorcycles that use an inductive alternator - BMW boxers come to mind - and they in theory would use more fuel. Though I would guess that the extra fuel consumed is so small as to be swamped by other factors such as the longer warm up cycle in cold weather, the greater air density in cold weather, etc.

 

[mcatrophy]

I guess I need someone to explain this to me. On another forum, a member posted that running heated gear lowers your miles per gallon.
I assumed that the stator is always putting out power, so there is no extra "drain" on the engine....thus fuel mileage would be not effected.

How would running more electrical accessories increase the load on the motor to effect fuel consumption?

This can be heavy. You are warned.

Gen I alternators run full current all the time. What isn't needed to charge the battery is dumped. This is where I confess ignorance, it depends how it is dumped. If the alternator is effectively short circuited, then little power is used (just self-heating of the alternator wires, which is happening anyway), so the load on the motor reduces while load-dumping occurs, more power is needed when the current is going to the battery/heated stuff. If the current is diverted into a resistor and is wasted, there is little difference in the power required from the motor.

Gen II alternators switch the current off when not charging the battery, so power definitely is reduced when not charging.

So, more power will be needed when running heated gear (or any other electrical load).

But, the power we are talking about is probably less than 100 watts (that would correspond to 8 amps). Lemme think (draws figure from 50 years ago) 746 watts=1 HP, so 100 watts is about 0.133 HP (1/7.5).

A typical figure for power used to move the motorcycle: 50mph needs something like 3.30x10^4W or 44HP (reference here).

So 100W isn't going to make a great deal of difference to your fuel consumption, something like 1/330.

 

[johnny80s]

Oh yeah a Good Friday thread!

 

[ionbeam]

This can be heavy. You are warned...something like 1/330.

Yum, that was tasty! :lol:

 

[Bearly Flying]

If you need your Electrics,( riding through a Blizzard or Heavy Rain) You aren't going to be worried about the minor loss of fuel milage.

 

[odot]

meh......................nerds

 

[Knifemaker]

Oh yeah a Good Friday thread!

Thanks

The bike in question is a Triumph Twin....which I thought used a fixed solid magnet spinning inside the stator windings to produce power. I'm new to the bike but all the info I've seen suggests that is the system that is used and its not an induction alternator.

 

[odot]

Hey, I got an idea.....run in a lower gear at the same speed and that increase in rotor rpm should make up for decrease in fuel economy.

 

[mcatrophy]

Oh yeah a Good Friday thread!

Thanks

The bike in question is a Triumph Twin....which I thought used a fixed solid magnet spinning inside the stator windings to produce power. I'm new to the bike but all the info I've seen suggests that is the system that is used and its not an induction alternator.

What vintage?

I had a '55 Tiger Cub about 100 years ago. It had the usual permanent magnet, three winding alternator (pretty much the same as on the FJR, but much lower power

). The light switch was a three-position one, Off, Side-lights, Headlight. When on Headlight, all three windings fed power to the rectifier (and hence to the battery), in the Side-light position, one winding was disconnected from the rectifier and short circuited, in the Off position, two windings were shorted leaving one to cope with the ignition power. There was no other control on battery charging.

From the riders hand-book:

(Click on image for larger view)



If your bike is like my cub, I'd be very wary of putting that much extra load on the battery. You dilithium crystals may be insufficient.

 

[Knifemaker]

Oh yeah a Good Friday thread!

Thanks

The bike in question is a Triumph Twin....which I thought used a fixed solid magnet spinning inside the stator windings to produce power. I'm new to the bike but all the info I've seen suggests that is the system that is used and its not an induction alternator.

What vintage?

I am not sure. The poster in question lists no model in his info box. I am assuming its a recent Bonneville model, SE, Scrambler, etc.

As it was posted in a Triumph Twins forum.

 

[mcatrophy]

...

I am not sure. The poster in question lists no model in his info box. I am assuming its a recent Bonneville model, SE, Scrambler, etc.

As it was posted in a Triumph Twins forum.

Ah, a "modern" twin, that should be ok, but I don't know its alternator rating. But the "logic" in my first post applies in terms of fuel consumption.

 

[Knifemaker]

...
I am not sure. The poster in question lists no model in his info box. I am assuming its a recent Bonneville model, SE, Scrambler, etc.

As it was posted in a Triumph Twins forum.

Ah, a "modern" twin, that should be ok, but I don't know its alternator rating. But the "logic" in my first post applies in terms of fuel consumption.

Keep in mind here my question was not "COULD you run heated gear and not increase fuel consumption .." But WHY would ANY extra electrical accessory decrease your fuel mileage on a motorcycle?

My thinking compared to jcyuhn post.... But the question as to WHAT causes an increased load on a motor when you add another electrical component? Does using more electrical power cause more resistance?

I see conflicting answers here. I've read about cars getting less mpg figures because of the "daylight headlights" but my thinking still says a bike using a magneto type stator always produces the same voltage output regardless of use.

Only if you begin using more amps than the system has to spare do you run into problems. How fuel economy is effected before this occours is the question.

 

[mcatrophy]

...But the question as to WHAT causes an increased load on a motor when you add another electrical component? Does using more electrical power cause more resistance?

...

To try to put things simply, the alternator (which Yamaha insists on calling a magneto) is driven by the engine, so the fuel consumption of the engine does depend on the load that the alternator presents to the engine.

The alternator, if electrically unloaded, would present only inertial loads on the engine plus a marginal mechanical loss in its bearings.

When the alternator is asked to deliver power by the regulator/rectifier, that power comes from the mechanical drive from the engine, plus some inefficiency due to winding resistance and eddy current losses in the metalwork. The efficiency won't vary much over the normal working range, and I guess the alternator would be something like 70% efficient, so if asked to provide 100 watts of electrical energy it would require 100 x 100/70 or about 143 watts of mechanical power, or if asked for 400 watts that would be about 400 x 100/70 or 571 watts.

At low speeds the alternator may not be capable of providing sufficient power, it's not producing enough voltage - the unloaded voltage is directly proportional to the speed.

The maximum current the alternator can ever deliver is limited by the winding inductance, this is a parameter that the original designer would be aware of in the specification. Somewhat counter-intuitively, you can't hurt the alternator by short-circuiting its windings, it limits its own current to this maximum value regardless of speed - inductive reactance is proportional to the inductance and the frequency, so as the speed increases the voltage increases but so does the frequency and hence the reactance, so the current remains the same. For a Gen II this current would be about 42 amps (14V, 590W, the specification is obviously after the regulator).

Equally counter-intuitively, when short-circuited, it is delivering no power (zero voltage at the short circuit, so whatever the current, the delivered power is zero). So, apart from the resistive and eddy-current losses, it's asking little mechanical power from the engine.

Charge a battery and/or supply the bike's electrical load, then the delivered voltage from the alternator is probably 16 or 17 volts (allowing for voltage drop in the regulator), now the power is significant, 16 x 42 x 100/70 = 960 mechanical watts demanded from the engine, or nearly 1.3 HP.

It is the regulator's job to limit the voltage at the battery to its nominal full charge value, a little over 14V. It does this by rapidly connecting the alternator to the battery or disconnecting it (and in the case of the Gen I shorting it to ground * ).

To sum up, whatever electrical power the alternator gives out is mirrored by the mechanical power from the engine (plus inefficiency losses), and so fuel burned.

Hope that gives some sort of explanation. I realise I've simplified, and left out bits, but that's the gist of it.

* By "ground" I mean the system 0V, battery negative, chassis, not true earth - I've seen some pedantic semantic arguments about this on this very forum

.

 

[Knifemaker]

I guess I'm still missing the "what" here. If you are going down the road at 60 mph, no wind nice flat road the bike is burning fuel to "crank" the crankshaft, which turns the gears, which turn the chain or shaft which turn the wheel to propel the bike down the road

To burn more fuel you need to add some kind of resistance to this. You can, for example, lightly drag a brake...this will cause you to start to slow down, so you compensate by twisting the throttle more to maintain the same speed.....thus burning more fuel.

If the end of the crank has a set of magnets on it, which are spinning inside a ring of coiled non magnetic copper wires, but not physically touching, how is the crank "slowed" by this?

To me it seems to be an open end system... If the voltage produced by the spinning magnets produces electrical power that is used or not used, what force is causing the crank to "work harder"?

If I take an electric fan and point it at a table, it's using X amout of power to run. If I put a tiny windmill on the table which powers a tiny generator, the fan is still using X amount of power. If that generator is powering a tiny light bulb, the fan still uses X amount of power whether the light is turned on or off. The fan does not slow down.

So, WHAT force is going to "slow down" that turning crankshaft when I turn on a light or plug an accessory in? To make the engine work harder and burn more fuel there had to be something physically trying to work against it.

That's the crux of my question here which still seems unanswered.

Sorry if I missed it.