Dyna Shopping

Reply Subscribe

Thread Tools

Search this Thread

May 6, 2009 | 07:56 AM

ColdCase

Thread Starter

Road Warrior

Joined: Apr 2009

Posts: 1,360

Likes: 4

Dyna Shopping

I'm new to the buying Harley experience. When a dealer does not have a rarer bike color/config in stock and offers to find one and get it sooner than later, but wants MSRP plus $330 shipping plus $350 prep for the effort, would that be reasonable?

Since it looks like MSRP with perhaps a $100 gift card is about as good as it gets out the door anyway (in New England), I am wondering if I wait till after the spring rush I would save the $680 immediate gratification penalty, or I am just delaying the inevitable. :-)

Doesn't seem to be anything currently on the used market I want.

May 6, 2009 | 08:01 AM

Sammi

Outstanding HDF Member

Joined: Mar 2006

Posts: 3,411

Likes: 22

From: Connecticut

That's no bargain.

Have you checked with other dealers to see if they can do better?

Dealers in my area requently swap bikes.

Sammi

May 6, 2009 | 08:37 AM

The Ground Truth

Cruiser

Joined: Feb 2009

Posts: 227

Likes: 0

From: TN

Places close to me in TN will give you 1500 either off the new bike or a gift card to the store. You can definitely do better.

May 6, 2009 | 08:51 AM

archergodwin

Elite HDF Member

Joined: Sep 2008

Posts: 4,440

Likes: 148

From: Northwest

For that kind of money, look outside of your area. Not sure how 'rare' you are looking for, or new or what, but consider getting a less expensive model, and working up to what you want. Especially if you don't want to look outside your area.

May 6, 2009 | 08:56 AM

SBRob

Seasoned HDF Member

Joined: Oct 2008

Posts: 7,645

Likes: 4,883

From: Missouri City,TX

It all depends the area you live in...the dealers in my area are still at MSRP (or more) plus shipping and set up.

I spent three days goin' back and forth with my local dealer, the best i could get was MSRP plus shipping and a set up fee w/ a $50.00 gift card, but I DID get more than wholesale book for my trade in.