Hey guys, question for a friend of mine.
He wants to ONLY do team challenge. What is the bare minimum of licenses he will need? I thought it was ASRA [which is now AMA] only. But this makes me think you would need the AMA and a CCS one.
QuoteAs announced in the early mailers, we have a new requirement for ASRA, Team Challenge & USGPRU. All of those competitors must have a current AMA Membership and have their card with them while registering at the track.
Please have your AMA Membership card as well as your CCS license and photo ID ready to show our registration clerks. (Your AMA Pro racing license is not your membership card.)
Need CCS license, and AMA membership.
Don't need to have an AMA race license.
Wait, what? It's called "ASRA Team Challenge" for a reason, no?
Rule book pg45: 9.3 Team Composition- Each team shall consist of a Team Captain, a CCS or ASRA licensed
rider who shall register the team with ASRA.
Not as of this year.
With the AMA sanctioning came the removal of the additional ASRA license fee.
Now, you will still need to check the "I wish to be licensed for ASRA competition" box on your CCS license form, or get assigned an ASRA number at registration.
You need a CCS license, and an AMA Membership to race in Team Challenge.
(AMA Membership is not an AMA Pro License. The Membership and Pro License are two completely different things, handled by two completely different organizations.)
There is no longer a fee for ASRA, but if you plan on racing in ASRA SuperStock, SportBike, or Thunderbike, you still need to sign up for an ASRA competition number.
AFAIK You do not need an ASRA competition number to race in Team Challenge, you just need a CCS license, and AMA Membership. As long as there aren't any other bikes with your number, you can run your CCS number.
Also, prior to the AMA sanction, you only needed a CCS license to race in Team Challenge, you didn't need an ASRA license or competition number. You needed a CCS license in order to also get an ASRA license, since the ASRA license wasn't a stand-alone.