Grip heater circuit

[Dads_FJ]

I'm adding grip heaters to my bike and drew up this circuit for high/low range.  Could use a 2nd opinion before I start sourcing parts etc...

[glfredrick]

How do you control the ultimate temp of the heaters.  I see you have two relays for high and low range tied into the system, but what causes them to actually be high and low range as sent to the heaters?

Otherwise, I see no difficulties with your circuit diagram.  Of course, I've never wired up grip heaters before either...  But, I do have some electrical experience and have stayed at least once in a Holiday Inn Express. 

[Dads_FJ]

How do you control the ultimate temp of the heaters.  I see you have two relays for high and low range tied into the system, but what causes them to actually be high and low range as sent to the heaters?

Otherwise, I see no difficulties with your circuit diagram.  Of course, I've never wired up grip heaters before either...  But, I do have some electrical experience and have stayed at least once in a Holiday Inn Express. 

On 'low' the grip heaters are wired in serial, which puts about 6-7 volts accross each one.  On 'high' they are connect in parallel, which would put about  12-14 volts accross each one.  The only thing I would change with this circuit would be to use a switched 12 volts from the igintion to power the relays, that way I won't forget them on when the bike is off.

[93fj1200]

http://faq.f650.com/FAQs/HotGripsInstallationFAQ.htm

[FJ Flyer]

Hot Grips use a resistor to go to 'low'.  I think a serial circuit will be way too hot.  When i tried repairing my widder glove harness i mistakingly wired the gloves in series.  Damn near gave me burrns almost instantly.  Rewired them in parallel and they worked fine, at least until the glove crapped out.

[SlowOldGuy]

Okay, I'm stupid on this stuff, so this seems like a good learning opportunity since we seem to have some experts on the subject.

What Chris says above is opposite to what is claimed in the first post, that serial resistance produces less heat than parallel resistance.  BTW, very clever and efficient electrical wiring, if it works.

According to my well honed misunderstanding of electricity, more current through a resistance = more heat/watts.

For a constant of 12 volts, the parallel resistors would equate to less equivalent resistance (R1*R2/(R1+R2)), thus allowing more current flow (and since P = I^2*R) more heat would be produced than the serial resistance which has greater equivalent resistance, less current and (supposedly in theory) less heat.

So can someone explain it to my tiny brain such that I can understand it?

DavidR.

[Dads_FJ]

Okay, I'm stupid on this stuff, so this seems like a good learning opportunity since we seem to have some experts on the subject.

What Chris says above is opposite to what is claimed in the first post, that serial resistance produces less heat than parallel resistance.  BTW, very clever and efficient electrical wiring, if it works.

According to my well honed misunderstanding of electricity, more current through a resistance = more heat/watts.

For a constant of 12 volts, the parallel resistors would equate to less equivalent resistance (R1*R2/(R1+R2)), thus allowing more current flow (and since P = I^2*R) more heat would be produced than the serial resistance which has greater equivalent resistance, less current and (supposedly in theory) less heat.

So can someone explain it to my tiny brain such that I can understand it?

DavidR.

I think you have it down perfectly Dave (and thanks for the complement!).  I don't understand why the gloves mentioned above produced more heat wired serial.  But according to my thinking (and yours and ohm's law), more resistance = less current, and there is more resistance when resistors are connected in series vs. parallel. 

[SlowOldGuy]

That's exactly what my EE 101 math is telling me also.  However, I've been known to jump to the wrong conclusion on occasion or interpret the obvious 180 degrees out from reality.

Just looking for an expert confirmation.

My confusion is that I always "thought" more resistance would equal produce more heat.  But the math says more resistance means LESS current flow which produces less I^2*R power dissipation (heat).

While this is backwards to my mechanical leaning brain, it kind of makes sense for other observations I've made on headlight bulbs.  Resistance of the hight "watt" bulbs tends to be less than the lower watt rated bulbs.  This matched the "grip" logic from above in that a lower resistance headlight filament allows more current to flow and thus produces more heat/light.

I think I'm still trying to convince myself.
DavidR.

[Dads_FJ]

Thinking of it in extrems can be helpfull.  Take an open circuit where the resistance is infinite - This will produce zero current.  So if Power (watts) = Volts x Amps then there will be Zero watts produced.  Or the opposite - take a toaster where the wire has very little resistance,  that will provide a great path for current to flow - using a lot of wattage, or power.

[carey]

Buy a set of Dual Star or similar grip heaters.  These are heating pads that fit under whatevewr grip you like.  The clutch and throttle sides are sized differently to give even heat because the throttle tube is less of a heat sink the the handlebar.  The cost about $35 plus shipping and draw about 36 Watts.  There is an unused switched power plug behind your side panel, one wire is brown.  On an 86, the plug was behind the left panel, but I think it's under the right panel on newer models.  I've installed  6 sets of these on various bikes and have tested them to 20 deg F.  Both of my street bikes have them installed now, and I have anither set on my workbench ready to install on my KTM 300.  No relays are required.

http://www.dual-star.com/index2/Rider/heated_grip_kit1.htm



http://www.casporttouring.com/cst/motorcycles/heated_grips.html

[Flynt]

So can someone explain it to my tiny brain such that I can understand it?

DavidR.

If your like me, the math helps...  

V=IR => for a constant voltage, I and R are inversely proportional...  formula is I =V/R

so a delta decrease Resistance (dR), you will have the inverse delta increase in Current (dI).  Now you can pretty easily conclude that shifting resistance into current pays off in heat due to the exponent (V=dI^2 * dR).

Practical example...  put a light build between your battery terminals and you get light.  Put a wire and you get burnt...   :big grin:

Hope that helps,

Frank

PS - somebody check that please...

[Flynt]

So can someone explain it to my tiny brain such that I can understand it?

DavidR.

(V=dI^2 * dR)

'V' should be heat... 

[FJ Flyer]

Ok, this was about two years ago.  Thats an equivelant to a millenium for my kid-fried brain cells.    I may have gotten backwards, i.e., supposed to be serial and I wired them in parallel.  I do know that they were damn hot.

Glad I only had to take one EE course.  Give me fluids and wings, any day.

[fj1289]

Give me fluids and wings, any day.

Like Hooters?!   

[SlowOldGuy]

Hey Chris,
Didn't mean to pick on you.  I was talking about this problem with EEs here at work and they claimed that (more resistance) = (more heat).

I argued the math (V=IR, P=I^2R, etc) and they just seemed to get confused.  At that point I figured it was a software problem, so I called the help desk.  :-)

DavidR.