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997 Turbo power kit
- Thread starter YvesD
- Start date
ORIGINAL: Lancerlot
No matter how much power is on offer, drivers always want more!
And yet, by far the best thing to add is - lightness.
Reduce the weight and you improve all the dynamics - performance, braking, steering and handling.
ORIGINAL: YvesD
Or even better, lightness and more power.
Presumably, that'll be the 997 GT2 then [8D]
Why aren't 480 geegees enough for you?
500+ for limited time on overboost? - wait until the tuners get their hands on that!
YvesD
New member
ORIGINAL: oliver
ORIGINAL: YvesD
Or even better, lightness and more power.
Presumably, that'll be the 997 GT2 then [8D]
Lets hope so !
ORIGINAL: YvesD
Kennyboy,
Where did you hear that ?
We were told at the 997 Register day out at Reading recently that peak power is not affected by the overboost but an additional 20bhp is developed accross the range.
Yves
I think you may mean 20 nm of torque.
YvesD
New member
ORIGINAL: Lancerlot
ORIGINAL: YvesD
Kennyboy,
Where did you hear that ?
We were told at the 997 Register day out at Reading recently that peak power is not affected by the overboost but an additional 20bhp is developed accross the range.
Yves
I think you may mean 20 nm of torque.
Nope, 60Nm of extra torque at its maximum is the official line and Porsche's man definately said 20bhp thoughout the range but no change in peak power.
It kind of makes sense because torque and power are related, can someone explain how you can increase torque by so much without increasing power at all ?
Yves
simonm
New member
Torque = Horsepower x 5252 / rpm
There are instances where this doesn't hold true e.g. when talking to your OPC salesman!
Or if the quoted 'power' is measured in PS:
1 BHP is 745.7 W whereas 1 PS is 'only' 735.6 W
So 1 PS is 1.014 BHP.
ORIGINAL: YvesD
Nope, 60Nm of extra torque at its maximum is the official line and Porsche's man definately said 20bhp thoughout the range but no change in peak power.
It kind of makes sense because torque and power are related, can someone explain how you can increase torque by so much without increasing power at all ?
Yves
I concur and with the courtesy of Wikipedia [8|]:
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Power is the work per unit time. However, time and rotational distance are related by the angular speed where each revolution results in the circumference of the circle being travelled by the force that is generating the torque. This means that torque that is causing the angular speed to increase is doing work and the generated power may be calculated as:
![33a0efe027e2d2d9719cf8780b8c4119.png](http://upload.wikimedia.org/math/3/3/a/33a0efe027e2d2d9719cf8780b8c4119.png)
Mathematically, the equation may be rearranged to compute torque for a given power output. However in practice there is no direct way to measure power whereas torque and angular speed can be measured directly.
Consistent units must be used. For metric SI units power is watts, torque is newton-metres and angular speed is radians per second (not rpm and not even revolutions per second).
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Conversion to other units
For different units of power, torque, or angular speed, a conversion factor must be inserted into the equation. For example, if the angular speed is measured in revolutions instead of radians, a conversion factor of 2Ï€ must be added because there are 2Ï€ radians in a revolution:
![12a4192b9d04e843879afe3ac587d8b6.png](http://upload.wikimedia.org/math/1/2/a/12a4192b9d04e843879afe3ac587d8b6.png)
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, foot-pounds (lbf·ft) for torque and rpm's (revolutions per minute) for angular speed. This results in the formula changing to:
![2e621cd9bcc2e222244b065aff8ad56b.png](http://upload.wikimedia.org/math/2/e/6/2e621cd9bcc2e222244b065aff8ad56b.png)
This conversion factor is approximate because the transcendental number π appears in it; a more precise value is 5252.113 122 032 55... It also changes with the definition of the horsepower, of course; for example, using the metric horsepower, it becomes ~5180.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius x angular speed. By definition, linear distance=linear speed x time=radius x angular speed x time.
By the definition of torque: torque=force x radius. We can rearrange this to determine force=torque/radius. These two values can be substituted into the definition of power:
![1af823f18b6c35d10a831c3e7dc53101.png](http://upload.wikimedia.org/math/1/a/f/1af823f18b6c35d10a831c3e7dc53101.png)
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2Ï€ in the above derivation to give:
![12a4192b9d04e843879afe3ac587d8b6.png](http://upload.wikimedia.org/math/1/2/a/12a4192b9d04e843879afe3ac587d8b6.png)
If torque is in lbf·ft and rotational speed in revolutions per minute, the above equation gives power in ft·lbf/min. The horsepower form of the equation is then derived by applying the conversion factor 33,000 ft·lbf/min per horsepower:
![cdc8e63f6e275a583fcfb6ed10a43cc7.png](http://upload.wikimedia.org/math/c/d/c/cdc8e63f6e275a583fcfb6ed10a43cc7.png)
Because 5252.113... = 33,000 / 2Ï€.
Anyway I'm sure you can have your 20bhp or 100 probably with a blown 911 with the tuners - I suspect after a drive you'll forget all about it! [
YvesD
New member
I worked my way through the Superbike scene until I had the ultimate 200mph balistic missile and you even get blase about that after a while ( 0-100mph in 5.5 secs and 150mph in 10 and a bit secs so I know where it all ends )
You pays your money and at that point the mostest thats available is what you want.
Off to therapy.
Yves
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