I would like to get smaller steel rims for my car in place of the 17 alloys, but I am not sure to what size I can go down to without causing issues with my rotors. I only have found 16.5, would like to go smaller can I do it..
Thank you anyone that can help me with this issue

 

Go to Tire Rack - Your performance experts for tires and wheels

Put in your car's information & go through the motions of buying a winter wheel/tire package. I think that might be the only way to see any available steel wheels, then check the size they give you. I bet it's 16".

 

well went to tire rack and really not helpful at all, kept me at 17 rims or bigger..Why is it so hard to get any accurate infornation on rims available for a vehicle, only give you same or bigger. Was told to come here to get help, humm....

 

OK I tried it; I got to 16 x 6.5 steel rims pretty quickly. Also, the basic Accord LX 4-door 4-cyl comes with steelies from the factory. If you have a V6 you might want to repeat that just in case your brakes are larger.

Select "Winter", then there's a choice between 16" or 17" packages. Then there's a link (not very big) that says something like "skip to wheels".

 

When you see the cost involved in changing wheels, tires, TPMSensors and the troublel you might cause with doing this-you more than likely will want to stay with your current tire and wheel package.

Welcome to the HAF..

WheelBrokerAng

 

Are you attempting to buy a second set of wheels and tires for your car for say Winter use, or do you want to move down for other reasons? If for Winter use then it makes a lot of sense to move to a 16 x 6.5 set of steelies and wrap them with winter tires, and as long as you keep the overall diameter of the tires the same there will be no issues with the TPMS.

FWIW, over at TireRack.com the default winter wheel/tire package for your car includes 215/60 R16 Bridgestone Blizzak WS70 Winter tires mounted on 16 x 6.5 steel wheels, new TPMS sensors, and a set of hub-caps. Total price is $845 plus taxes, shipping and handling.