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...
(http://i208.photobucket.com/albums/bb37/campsimonette/grips.jpg)
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.
Quote from: glfredrick on February 24, 2012, 08:56:17 AM
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.
http://faq.f650.com/FAQs/HotGripsInstallationFAQ.htm (http://faq.f650.com/FAQs/HotGripsInstallationFAQ.htm)
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.
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.
Quote from: SlowOldGuy on February 24, 2012, 02:04:21 PM
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.
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.
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.
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.dual-star.com/index2/Rider/heated_grip_kit1.htm)
http://www.casporttouring.com/cst/motorcycles/heated_grips.html (http://www.casporttouring.com/cst/motorcycles/heated_grips.html)
Quote from: SlowOldGuy on February 24, 2012, 02:04:21 PM
So can someone explain it to my tiny brain such that I can understand it?
DavidR.
If your like me, the math helps... :mail1:
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...
Quote from: Flynt on February 24, 2012, 09:26:14 PM
Quote from: SlowOldGuy on February 24, 2012, 02:04:21 PM
So can someone explain it to my tiny brain such that I can understand it?
DavidR.
(V=dI^2 * dR)
'V' should be heat...
Ok, this was about two years ago. Thats an equivelant to a millenium for my kid-fried brain cells. :crazy: 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.
Quote from: FJ Flyer on February 25, 2012, 08:08:04 AM
Give me fluids and wings, any day.
Like Hooters?! :drinks:
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.
Quote from: SlowOldGuy on February 27, 2012, 10:03:10 AM
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.
Adding more resistance would mean more heat
in the resistor, but less in the gloves, yes?
I would suggest wiring everything up on the bench and testing it before taking all the time on the bike.
Hey dads fj.
the wiring looks good .
for the lectrical challenged keep it simple . more resistance =less amps ....... less resistance =higher amps..... higher amps = mo' heat.
Quote from: rktmanfj on February 27, 2012, 10:06:29 AM
Adding more resistance would mean more heat in the resistor, but less in the gloves, yes?
Randy,
The grip heating elements ARE the resistors.
DavidR.
Quote from: SlowOldGuy on February 27, 2012, 02:36:02 PM
Quote from: rktmanfj on February 27, 2012, 10:06:29 AM
Adding more resistance would mean more heat in the resistor, but less in the gloves, yes?
Randy,
The grip heating elements ARE the resistors.
DavidR.
Do some setups not use switched resistors to adjust the heat?
Quote from: rktmanfj on February 27, 2012, 03:31:56 PM
Quote from: SlowOldGuy on February 27, 2012, 02:36:02 PM
Quote from: rktmanfj on February 27, 2012, 10:06:29 AM
Adding more resistance would mean more heat in the resistor, but less in the gloves, yes?
Randy,
The grip heating elements ARE the resistors.
DavidR.
Do some setups not use switched resistors to adjust the heat?
This model uses dual heating elements http://pollyheaters.com/ph400.htm (http://pollyheaters.com/ph400.htm) However I already ordered the ph300 single element. My plan is to hook them up without drilling holes etc... before I mount them permanently.
Oxford heated grips, with heat control would save you any gue (popcorn)ss work
Quote from: fj11.5 on February 27, 2012, 08:39:51 PM
Oxford heated grips, with heat control would save you any gue (popcorn)ss work
+1. I've been thinking the same thing throughout this entire thread. Yeah, you could build your own and save a few bucks, but it IS only a "few bucks". Oxfords can be had for like 35 bucks....
Exactly , I paid $120 for the original pair , the second gen ones with heat control were $70 a few years ago, so as you said , wouldnt be saving much, (popcorn)
Quote from: fj11.5 on February 27, 2012, 09:37:37 PM
Exactly , I paid $120 for the original pair , the second gen ones with heat control were $70 a few years ago, so as you said , wouldnt be saving much, (popcorn)
Pretty much the story of my life, trying to save a few bucks but usually costs more in the long run. Usually because I'll dick around with a project (having fun at the same time though), but end up buying something like Oxford grips recommended here! You'd think I'd learn. I bought the Pollys for $21 to my door, so I figure I'm not too far in the hole yet. Still it's all fun...
Nothing wrong with making your own stuff , especially if it works, never thought of making heated grips (popcorn)
Quote from: Dan Filetti on February 27, 2012, 08:57:00 PM
Quote from: fj11.5 on February 27, 2012, 08:39:51 PM
Oxford heated grips, with heat control would save you any gue (popcorn)ss work
+1. I've been thinking the same thing throughout this entire thread. Yeah, you could build your own and save a few bucks, but it IS only a "few bucks". Oxfords can be had for like 35 bucks....
I miss spoke, I did not get the Oxfords, I got a set like these:
http://www.ebay.com/itm/YAMAHA-HONDA-KAWASAKI-SUZUKI-GRIP-HEATERS-/320770949359?pt=Motorcycles_Parts_Accessories&vxp=mtr&hash=item4aaf7040ef (http://www.ebay.com/itm/YAMAHA-HONDA-KAWASAKI-SUZUKI-GRIP-HEATERS-/320770949359?pt=Motorcycles_Parts_Accessories&vxp=mtr&hash=item4aaf7040ef)
I see even cheaper ones on eBay as well.
FYI,
Dan
My Oxford grips are coming up for their 4th winter and have about 45,000 k's on them. If they died tomorrow I'd still reckon they were fantastic value. And if they did die tomorrow I'd be at the shop for a new pair before the old ones had even cooled.
Noel
Dan,
That link was for Symtec grip heaters. I've installed many sets of them. There was a set on my FJ when I sold it. Garth, are they still working?
Carey
Quote from: carey on February 28, 2012, 06:25:11 PM
Dan,
That link was for Symtec grip heaters. I've installed many sets of them. There was a set on my FJ when I sold it. Garth, are they still working?
Carey
Yep they rock. I just bought another set, dual star branded but same thing.
+2 for the oxfords. I put a set on my FJ and a set on Donna's Ninja 650. A little over $100 each canadian right from the local bike shop. Great investment. Donna's are newer and have a different controller for them. They seem to put out a bit more heat. I bought a third set for my brother's ZX14 but haven't installed them yet. Maybe after all this spring's FJ projects....
Derek
I bought a couple of sets of grip heaters from a vendor to try on my bike. The are wired the same as the original post with a toggle switch for high & low.
They sell for $29.98 if anyone is interested. I have a set on my bike, but I have not ridden much this year to try them out.
If interested let me know,
Randy - RPM
Quote from: Dads_FJ on February 27, 2012, 03:36:21 PM
Quote from: rktmanfj on February 27, 2012, 03:31:56 PM
Quote from: SlowOldGuy on February 27, 2012, 02:36:02 PM
Quote from: rktmanfj on February 27, 2012, 10:06:29 AM
Adding more resistance would mean more heat in the resistor, but less in the gloves, yes?
Randy,
The grip heating elements ARE the resistors.
DavidR.
Do some setups not use switched resistors to adjust the heat?
Yep, the Hot Grips have a resistor in the system to switch between High and Low settings.