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Koenigsegg Regera - acceleration

  1. Mar 9, 2015 #1
    hi guys
    did you noticed announcement of new koenigsegg regera?
    1500hp, two electric motors driving wheels, and one connected to engine. no transmission, only hydraulic clutch and final drive.
    interesting thing is, that they said the engine engages at ~50kmh, and at lower speeds, electricity is moving the vehicle. they also published a graph of power and torque figures of whole propulsion system. 0-60mph at 2.8sec, 0-250mph at sub 20 seconds.

    my problem is this. if you look at the combined torque figure ( at 1000rpm) there is around 600Nm of electric torque. we can assume, that that is torque of two electric motors. there is final drive (2.85:1) between engine and motors, so that two motors would produce aprox (600nm*2.85) 1710Nm going to the wheels. and that is actually very very little. 2l petrol engine with transmission (lets say 4:1 for first gear and 4:1 final drive) sending to the wheels like (200nm*4*4) 3200Nm

    I put it all (weight, aprox torque curve, gear, tyres, aero coefficient...) in my excel sheet which can calculate thinks like this and with this torque, it calculated that 0-50kmh would take almost 5 seconds and 0-100kmh something like 7seconds that is more than twice more than official time. other acceleration figures (90-150mph and 30-250mph) it calculated very close to official figures.

    so help me here. did I make somewhere a mistake? because unless the published power/torque curves are wrong, I can't understand how it could achieve that 0-60mph acceleration time.
  2. jcsd
  3. Mar 9, 2015 #2


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    Combined torque is given as 900 Nm for 1000 rpm. I don't know the size of the wheels so it is tricky to convert that to a speed.

    Accelerating to 50 km/h takes about 150 kJ, at the given power value of 110 kW (at 1000 rpm) this would just take a bit over 1 second - the wheels might slip if you try that.
  4. Mar 9, 2015 #3
    we know that it reaches 400kmh at 8250rpm with final gear of 2.85:1. from that we can calculate wheel radius of aprox 0.367m
    thing is, that until 45-50kmh the engine should be disengaged from powering the wheels. so we have only 600Nm of electric motors, and the power output scales with revs. 600Nm at 1000rpm produces 62kW, at 500rpm it is 31kW etc
  5. Mar 10, 2015 #4

    jack action

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    I think they have it wrong in their explanation in this article.

    The electric motor/engine combination is used at any speed in conjunction with the 2 other electric motors. When they talk about an «hydraulic coupling that acts like a clutch, [...] This sounds similar in concept to a torque converter», it is in fact a torque converter. A torque converter also multiplies torque when it is slipping at low rpm. The typical multiplier effect is 1.8:1 to 2.5:1 and can be up to 5:1. Let's assume it is 2:1 for now.

    The engine looks like it can produce about 500 N.m of torque at a stall rpm of about 1600 rpm. Let's assume the electric motor makes 300 N.m like the other two. That is a combined torque of 800 N.m. After the torque converter, it becomes 1600 N.m and after the gear reducer it becomes 4560 N.m.

    Then you add the 2 other motors producing an extra 600 N.m for a total of 5160 N.m.

    The tractive force with a 0.367 m tire radius is 14 060 N.

    The car weights 3589 lb (1628 kg). Add a 70 kg driver and you get a total of 1698 kg. It is pretty safe to assume the car has a 50/50 weight distribution, so that is 849 kg on the rear wheels or a normal force of 8329 N acting on the rear tires.

    The tire coefficient of friction to support such a tractive force needs to be 14 060 / 8329 = 1.7. That is about the largest tire coefficient of friction that you can find on a tire (apart from drag racing tires).

    I may have overestimated stall rpm, torque multiplier and/or some torque rating, but then again 1.7 is also a pretty high number for a CoF.
  6. Mar 11, 2015 #5
    yes, explanation that hydraulic clutch is actually torque converter and the engine is powering the car from (almost) standstill would make lots of sense, and solve this problem.
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