Durango fog lights wattage

From : peter

Q: tbone wrote nothing fuzzy about it you might want to pull out the 4th grade math book the one with dick and jane and spot on it and do some reading from what i am reading from you here i can see that dick and jane is about as far as you got in your education. then ask yourself is this a true statement 0.03 / 0.02 = 1.5 yes or no sure it does but the real question is what does .03 and .02 stand for on their own and the answer is nothing at all. and if they stand for nothing as is what in the hell does the 1.5 you get by dividing them stand for i think that you will find the answer is the same as the last one nothing. for this reason alone it either shows a failure in your logic or that you are a graduate in the gwb fuzzy math academy. your calculation is nothing more than a percentage of a percentage and when you do that you lose just about all valid meaning. for the momory-impaired such as yourself 0.03 stands for amount of dirt 3 % allowed to pass by a k&n and 0.02 2% is the amount passed by an oem filter therefore a k&n pases 150 % or 50 % more dirt than an oem get up to speed will ya once again your logic fails. as i said you are taking a percentage of percentages which has no valid meaning in itself. then on top of that your wording is wrong. if we were to use your logic it would let in 150% as much dirt as a paper filter not more than a paper one. if it was 150% more then it would be allowing 2 .5 times the dirt as paper and this is not true. it lets in up to 1.5 times as much dirt but when you look at how little dirt the paper element lets in 1.5 times just about nothing is still just about nothing. any way that you want to look at it from a valid baseline the k&n will allow up to 1% more dirt in than a high quality paper filter big deal. not trying to start a new argument here but why use it then if it isnt any different than a paper one .

Replies:

From : peter

how about an old fashioned gas war that forces the price of ethyl from $0.269 to $0.169 per gallon. id vote for that war. decimal points are nasty dirty little things arent they ;^ mike .

From : trey

can i put in aftermarket fog light assemblies rated at 55w in 99 durango will relay & wiring take it looks like the fog lights are run though a relay from the factory. its a 20a circut to the fog lights. volts x amps = watts correct i think thats right low end 10v x 20a = 200w high end 14v x 20a = 280w running two 55w laps would be 110w on paper that gives you a 90w cussion. here is a link to the page with the fog lights from my chilton repair manual. http//www.socalrider.net/2005/fogs.jpg think that pretty much says it. thanks! 20a fuse & relay should take it no problem! peter .

From : bob

can i put in aftermarket fog light assemblies rated at 55w in 99 durango will relay & wiring take it looks like the fog lights are run though a relay from the factory. its a 20a circut to the fog lights. volts x amps = watts correct i think thats right low end 10v x 20a = 200w high end 14v x 20a = 280w running two 55w laps would be 110w on paper that gives you a 90w cussion. here is a link to the page with the fog lights from my chilton repair manual. http//www.socalrider.net/2005/fogs.jpg think that pretty much says it. .