I had a 95 u.k turbo at 208 bhp, then a stint in an ibiza 16v gti at 150 bhp,now i`m back in a scoob and have a 97 wrx which even at conservative estimates rates at 260 bhp,this is 25%more power than my old u.k car, but why doesn`t it feel 25% quicker?I mean, my ibiza felt 25% slower.

 

top power output is just that, the max that can be screwed out of an engine, cars feel quick when they have torque (the flatter the torque curve and the higher, the better)

 

Moz

I assume your WRX is not an STI.

Your UK car would have put out about 225-230bhp. The WRX does not have loads more torque than your old UK car. 10% maybe.
Therefore I would not expect it to feel 25% faster.

Jonathan

waiting to be correted

 

The BHP value is a calculation of the torque output from your car at given rpm. 2 cars producing 220ft/lbs at 4000rpm wil produce the same BHP at these revs (even if one is a 2.0 Turbo and the other is a 5.0 V8). The reason you do not feel a large difference is because the standard turbo produces 220-230ft/lbs of torque compared to 250-260 for the WRX - around 10-15% difference. The other reason is that the standard turbo starts spinning about 1000rpm earlier, and feels more tourquey low down. The reason for the large difference in BHP (230 vs 280) is that WRX/STI's will rev for an extra 1000rpm.

 

After thinking long and hard about it I think of it this way. The higher the BHP figure, the quicker the engine can accelerate the torque output of the engine.

Weight then plays a big part in the equation, cause the heavier the vehicle the slower it will accelerate for a given torque and power output.

e.g. Turbo diesel engines have high torque figures but low BHP in comparison. Therefore they can pull like the blazes up hills but dont accelerate like crazy because of the low BHP figure and short power band.

Motorbikes have high BHP but small torque, but go like blazes cause the weight is small in comparison.

Of course it is important also where maximum torque and maximum horse power occur, because this has a direct effect on the accelerative force.

Hope this helps
Greg