2014-2020 Formula One 1.6l V6 turbo engine formula

[machin]

That single gear would have to be the gear top speed is reach, like a 7th gear. It can still lose out to constant power engines with more than 1 gear, since those gears will have a torque advantage.

Eh? That doesn't make sense; As Xpensive says above: the relationship between torque and power is inversely proportional. An engine making ~600BHP at 12000rpm will produce 262.5lbft of torque. Lets say it has a diff ratio such that this = 200mph. At 6000rpm (100mph) it also makes 600bhp, but now the torque it produces is 525lbft. At 3000rpm (50mph) it still makes 600bhp but now torque is 1050lbft.

If we added a 4:1 gear ratio in there so that the 262.5 lbft made at 12000rpm is increased by 4 we have 1050lbft, but the speed is 4 times slower, i.e. the gearbox output speed is 12000/4 = 3000rpm. 1050lbft of torque at 3000rpm = 600bhp... so we haven't added anything!

I still think that such an engine would be really heavy, so I bet they still go with constant power over a smaller rev range and have gears, but just wanted to point out that a constant power engine's performance is not increased by adding additional gears....

[PNSD]

Im not a mech kind of guy... so could this see a return to 6 speed gearbox's? Or would 7 still be the ideal choice?

[xpensive]

As power is Force times speed, while Force is accelleration is times mass, your accelleration will always be:

Power / (speed * mass)

There is no torque in that equation whatsoever, provided that you can have full power from the word go of course.

[JohnsonsEvilTwin]

As power is Force times speed, while Force is accelleration is times mass, your accelleration will always be:

Power /(speed * mass)

There is no torque in that equation whatsoever, provided that you can have full power from the word go of course.

The very ethos behind my little S2000

[autogyro]

Gearboxes do a great deal more than just join the engine to the drive wheels.
In this case it is worth remembering that the lower gears multiply torque at the wheels and also reduce the wheel speed.
Very low wheel speed from standstill means no wheelspin with very high torque, equals very rapid acceleration. Limitation is the mass of the car against the strength of the geartrain. Oh yes, and available traction.

It is why it is possible to get a 0 to 60 time of under 4 seconds with 150 bhp when a one gear dragster with 2000bhp will sit spinning its wheels.

Sorry JET, I had a Fiat 500 with a mid mounted Lotus Twin cam engine that would easily beat your car to 100 without trying, no matter what power it has.

[NewtonMeter]

As power is Force times speed, while Force is accelleration is times mass, your accelleration will always be:

Power / (speed * mass)

There is no torque in that equation whatsoever, provided that you can have full power from the word go of course.

In you equation, which I interpret as:

a(t) = P/ (v(t) * M),

torque is implied by your splitting of Force into mass and acceleration. It can also be seen in your very equation which, at face value, features acceleration as the result, but no force. But the only way to induce acceleration is by a net force...so there HAS to be a force which provides the driving force for the acceleration. That's Newton 2 for you right there. And the only place that force can originate from is engine torque, which then goes through the gearbox (and gets modified by gear ratios), through the diff (and gets modified again by the final drive ratio), through the wheels and is then translated as force at the contact patch of the wheels which accelerates the car forward.

So, in the equation you propose, the force to drive car forward would come from the power term. Which would be broken up into torque and angular velocity (which is proportional to RPM) at the wheels . Constant full power would then imply that:

P(t) = P = T(t) * w(t). But it would still mean that T(t) would have to diminish in proportion as w(t) increases. Therefore giving an equation as follows:

a(t) = (T(t) * w(t))/ (v(t) * M),

and there's your torque.

Edit: Basically, your equation is really more of relationship than 'n function.

[machin]

Basically, Power = Torque x Speed

If power is constant, and speed decreases, torque has to increase. You cannot improve the rate of work done by a machine which is truly constant power by introducing a gear ratio.

[autogyro]

Basically, Power = Torque x Speed

If power is constant, and speed decreases, torque has to increase. You cannot improve the rate of work done by a machine which is truly constant power by introducing a gear ratio.

It is far from that simple.
True you cannot improve the rate of 'potential' work by introducing a gear ratio.
You can however make available work at that gear ratio possible.

[machin]

Auto; explain please how you can increase the rate of energy output of a theoretical machine that has the same power output regardless of output speed by adding a gearbox?

[autogyro]

I did not say you can increase the rate of 'potential' energy output from a 'theoretical' machine.
I said you can make energy available using gear ratios in reality, when it would not be without them.

[machin]

OK, so you're ignoring the topic we're talking about here which is constant power machines and whether gearboxes improve them (which they don't). I think we're all aware that you can improve a machine with a very "peaky" power output by adding gears so that you can have that power output available at different speeds.

[autogyro]

OK, so you're ignoring the topic we're talking about here which is constant power machines and whether gearboxes improve them (which they don't). I think we're all aware that you can improve a machine with a very "peaky" power output by adding gears so that you can have that power output available at different speeds.

That is not what I am saying at all.
It has nothing to do with power delivery.
2 bhp will give you huge acceleration from standstill to 10mph if the gear ratio is low enough, no matter how constant the power from the machine delivering it.

Gear ratio ranges will always improve acceleration in gear driven vehicles.

[machin]

They will only improve the acceleration of a vehicle which only produces peak power at a small range of speeds. They won't improve the acceleration of a vehicle which produces the same power at all speeds. Otherwise you're braking the laws of physics.

[autogyro]

They will only improve the acceleration of a vehicle which only produces peak power at a small range of speeds. They won't improve the acceleration of a vehicle which produces the same power at all speeds. Otherwise you're braking the rules of physics.

Sorry Machin, you are still stateing 'theoretical' potential.
It does not matter if you apply the same power at 4000rpm or 12,000rpm with the same power available, in reality using a one to one ratio and 650 bhp, the result will just be wheelspin and you will go nowhere.

[WhiteBlue]

I think the discussion about having or not having gearboxes is theoretical and not practical. The FiA will not suddenly let you off with an F1 car that has no reverse gear. So a gearbox will always be required. The question is will it be beneficial to reduce the number of gears and the gear ratios drastically? You also have to consider that the gearbox may have to have a second input shaft from the turbine.