T5 performance - Page 2

Sep 3rd, 2002, 08:27

Judging by the fact we saw you drive past as we came out of the petrol station on Saturday, proceeded to travel 'as fast as the conditions allow' and yet simply could not find you would suggest Mr. W might have accidently got his foot caught on that little pedal on the right hand side :)

Sep 4th, 2002, 07:31

Just the spirit..
:)

davebslater(uk) · Sep 4th, 2002, 10:51

just a note on aerodynamics , all forces acting on the car increase by the square of speed , drag , lift , yaw so everything is concerned . you can take most things to 180mph with 500bhp , it would be faster if it had less drag - also that would normally mean less lift .

Sep 4th, 2002, 11:28

The power required to overcome rolling resistance is linearly proportion to the speed, but the power required to overcome aerodynamic resistance is proportional to the cube of speed... so to go twice the speed you need approximately 8 times the power <BLIMEY>.

So if the Volvo has a current max speed of 160mph and 240bhp, it would need about 370bhp-ish (+stainproof underwear) to make it to 200mph. Chip anyone?

Not sure about lift etc... I was in the pub during that lecture ;)

Andy

davebslater(uk) · Sep 4th, 2002, 16:49

sorry , square of the speed !! not cube

cd = Drag force in newtons / 1/2 * Density * Velocity ^2 * Area

you were obviously in the pub for that one aswell

Aerodynamics played a big part in my major project , so watch out

Sep 5th, 2002, 09:28

No, I think you will find I'm right :9

Since:

Power = Force x Velocity

And as you correctly state the FORCE required to overcome aerodynamic drag goes up with the square of velocity => the POWER to overcome aerodynamic drag goes up as the cube of velocity.

So there you go, every day is a school day!

Andy

rstrang · Sep 5th, 2002, 10:50

Hi Andy,

At the risk of getting flamed...:-)

Isn't rolling resistance a constant?.

Richard

Sep 6th, 2002, 11:45

No flames here :)

Rolling resistance is normally approximated to a constant force, which means (as above) the power to overcome rolling resistance goes up linearly with speed. I've never seen any real data for tyre rolling resistance (I'm an engine man really) so I'm not sure how good an approximation it is.

If you go here, there are applets for working out your CdA and rolling resistance:

http://www.race-technology.com/WebPa..._utilities.htm

Andy

rstrang · Sep 6th, 2002, 12:07

I was looking at this site:

http://pirelli.pearlcomm.com/pneumat...ining/roll.htm

Shows how bad Cross-Plys were, I guess most of that energy gets
converted into heat.

Richard

Sep 3rd, 2002, 08:27	#11
alexzoutsos Guest Location:	RE: T5 performance Judging by the fact we saw you drive past as we came out of the petrol station on Saturday, proceeded to travel 'as fast as the conditions allow' and yet simply could not find you would suggest Mr. W might have accidently got his foot caught on that little pedal on the right hand side :)

Sep 4th, 2002, 07:31	#12
gareth Guest Location:	RE: T5 performance Just the spirit.. :)

Sep 4th, 2002, 10:51	#13
davebslater(uk) redbollock guru Last Online: Jun 30th, 2016 23:54 Join Date: Jan 1970 Location: lincoln	RE: T5 performance just a note on aerodynamics , all forces acting on the car increase by the square of speed , drag , lift , yaw so everything is concerned . you can take most things to 180mph with 500bhp , it would be faster if it had less drag - also that would normally mean less lift . __________________ www.dbsmotorsport.co.uk

Sep 4th, 2002, 11:28	#14
Andy D Guest Location:	RE: T5 performance The power required to overcome rolling resistance is linearly proportion to the speed, but the power required to overcome aerodynamic resistance is proportional to the cube of speed... so to go twice the speed you need approximately 8 times the power <BLIMEY>. So if the Volvo has a current max speed of 160mph and 240bhp, it would need about 370bhp-ish (+stainproof underwear) to make it to 200mph. Chip anyone? Not sure about lift etc... I was in the pub during that lecture ;) Andy

Sep 4th, 2002, 16:49	#15
davebslater(uk) redbollock guru Last Online: Jun 30th, 2016 23:54 Join Date: Jan 1970 Location: lincoln	RE: T5 performance sorry , square of the speed !! not cube cd = Drag force in newtons / 1/2 * Density * Velocity ^2 * Area you were obviously in the pub for that one aswell Aerodynamics played a big part in my major project , so watch out __________________ www.dbsmotorsport.co.uk

Sep 5th, 2002, 09:28	#16
Andy D Guest Location:	RE: T5 performance No, I think you will find I'm right :9 Since: Power = Force x Velocity And as you correctly state the FORCE required to overcome aerodynamic drag goes up with the square of velocity => the POWER to overcome aerodynamic drag goes up as the cube of velocity. So there you go, every day is a school day! Andy

Sep 5th, 2002, 10:50	#17
rstrang Junior Member Last Online: Aug 17th, 2007 14:53 Join Date: Dec 2001 Location: Reading	RE: T5 performance Hi Andy, At the risk of getting flamed...:-) Isn't rolling resistance a constant?. Richard

Sep 6th, 2002, 11:45	#18
Andy D Guest Location:	RE: T5 performance No flames here :) Rolling resistance is normally approximated to a constant force, which means (as above) the power to overcome rolling resistance goes up linearly with speed. I've never seen any real data for tyre rolling resistance (I'm an engine man really) so I'm not sure how good an approximation it is. If you go here, there are applets for working out your CdA and rolling resistance: http://www.race-technology.com/WebPa..._utilities.htm Andy

Sep 6th, 2002, 12:07	#19
rstrang Junior Member Last Online: Aug 17th, 2007 14:53 Join Date: Dec 2001 Location: Reading	RE: T5 performance I was looking at this site: http://pirelli.pearlcomm.com/pneumat...ining/roll.htm Shows how bad Cross-Plys were, I guess most of that energy gets converted into heat. Richard

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