Thread: Motor bigbikes,250cc and more?

Hello Forum-Friends,

I was wondering if there is someone here who can help me finding a place where they sell 'real' motorbikes. With 'real' bikes, I mean 250cc or more.
I have searched the archives...searched using google, yahoo, and livesearch, no results
Only one result for renting...and not alot of choice there, only two bikes :(.
As I have browsed thru the archives here, I noticed that there are some bike enthousiast here....Help me plz... :D

friendly greetings......Gilbert.

In the North you should talk to Pak Logo.

Follow the road down from Panji until Jalan Raya. Half way(about) between Jalan Tengah and Jala Raya, you see "Gang Harley". (Sekretariat Harley Davidson Club)

He has a Suziki Bandit now, 400cc (20jt), and some Harley's. Even one for 85 jt !

I bicycle with him, his son and even his daughter 8)

Careful Bert,

you might get banned for breaking the forum rules, no ads, not even for bicycle buddies! :wink:

Sorry, couldn't resist :oops: 8)

Its an answer to a question IMHO :mrgreen:

And here he is, next to the handsome biker.

guess downhill you both should be faster than the rest -eh? 8)

Thanks Bert,
second time u helped me out already :oops: .
Dentist was pretty good :) .
And now this tip, I'll be going down gang harley tomorrow,to check out that bandit.
With bandit I dont mean pak logo(hahaha)but the bike 8) .

Ow yeah Torsten..I know ur only joking, but I looked at the forum rules...It says something like...what did you buy,where did you buy it, and how much did you pay for it..So I figured this was the place to ask..Instead of on the expat forum :)

Thanks again Bert,
Friendly greetings, Gilbert.

Hows your coughing, Thor?

Thorston,

You just flunked Physics 101. :) Newton's 2nd Law: the acceleration of an object resulting from a net force is directly proportional to the magnitude of the net force and inversely proportional to the mass of the object, i.e. a = F/m :lol: :lol:

I don't think so, Jim. This is the most elementary physics you can get at. It seems to go against common sense, which is quite often wrong. Thorsten wrote

guess downhill you both should be faster than the rest -eh?

Wrong. As I wrote above, a = F/m. The force exerted in this case by the earth on Bert (and reciprocally, Bert on the earth) is F = G m(earth) x m (Bert)/R*2
G = Gravitational constant
R = separation of the two bodies

So, the acceleration of Bert falling down the incline is

a = F (see above) / m (Bert), which is independent from the mass of Bert, and all bodies will reach the same velocities at any point along the trajectory (neglecting the air resistance)

Don't you remember the famous experiment by Galileo at the tower of pisa, which demonstrated this fact? :)