rec.autos.simulators

>So far I have run 4.36 in TF and 4.74 in FC.  I really don't think it's
>possible to go much quicker than that (maybe 2 or 3 hundreths).  I haven't
>spent quite as much time on the Funny Cars though, so maybe there is more
>left in it.

>    Coh, 4.26 for TF and 4.69 FC are the best I have seen, although
>I have only been able to get 4.29 and 4.71 myself!
>>So far I have run 4.36 in TF and 4.74 in FC.  I really don't think it's

>>>So far I have run 4.36 in TF and 4.74 in FC.  I really don't think it's

>What are you using for steering?

> > Every road car can generate a resultant force on the occupants in
> > excess of 1G. Sitting at rest or at constant velocity, there's a 1G
> > force (down). If the car brakes on a level surface at 0.8G (most any
> > road car can achieve this), the resultant force is (sqrt (+ 1 (* 0.8
> > 0.8))) or 1.28 G. For a 1 G lateral acceleration on a level surface,
> > (sqrt 2) or 1.4 G is the resultant force on an occupant.

> True again (and it gets worse if you consider 3d force vectors) but
> I still reckon most people will not notice the difference.  If you
> want it totally accurate I guess you need a very fast acting
> centrifuge :-)
> BTW. according to my sums, 1G down plus 0.8G back gives 1.34G at an
> angle of 51 deg from the horizontal..... ....just trying to get my
> own back for the Velocity comment :-)

>> > Every road car can generate a resultant force on the occupants in
>> > excess of 1G. Sitting at rest or at constant velocity, there's a 1G
>> > force (down). If the car brakes on a level surface at 0.8G (most any
>> > road car can achieve this), the resultant force is (sqrt (+ 1 (* 0.8
>> > 0.8))) or 1.28 G. For a 1 G lateral acceleration on a level surface,
>> > (sqrt 2) or 1.4 G is the resultant force on an occupant.

>> True again (and it gets worse if you consider 3d force vectors) but
>> I still reckon most people will not notice the difference.  If you
>> want it totally accurate I guess you need a very fast acting
>> centrifuge :-)

>> BTW. according to my sums, 1G down plus 0.8G back gives 1.34G at an
>> angle of 51 deg from the horizontal..... ....just trying to get my
>> own back for the Velocity comment :-)

>Don't know. I got roughly 51.3 degrees at 1.28 G. (The 51.3 is a
>little sketchy as I didn't use a calc with cos-1 implemented, so I did
>a quick Newton's approximation.)

>But, the magnitude of the resultant vector will clearly be (from
>Pythagorean theorem) (sqrt (+ (* 1 1) (* 0.8 0.8))), which is 1.28...

>Sorry,
>---Jim