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Old 6th February 2010, 01:24   #31
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power is simply a product of torque and RPM, isn't it? At low RPM if you are not producing enough torque, you are not producing enough power.

Power is simply a product of torque and RPM, isn't it? At low RPM if you are not producing enough torque, you are not producing enough power.


In other words, what you are saying is two engine "rated" at same power produce different power at low RPMS, and the one that produce more power wins.

that way the discussion will be simplified.


That's why I said power-rpm curve needs to be accounted for.
Thats what i've been saying.

Horsepower = torque (lbft) x rpm / 5252.

So an engine producing 686hp@8000 and an engine producing 686hp@4000 with have completely diff torque/power charateristics and will perfom differently. Any engine producing the same max power at lower revs has to produce more torque.



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That was exactly my point. I had considered an idealised situation, I mean, no drag force or anything. Even then the required power was hovering around 690 HP. In the real world, there are so many -ve factors like friction, drag, power loss etc. which will increase the engine power required.
So you basically want a 1600kg car to accelerate at the rate of 8m/s^2 without "friction"?




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First thing : Atom is not 1600kgs but an average car.

2. The discussion here is on modifying an average car whereas McLaren, Bugatti and F1 are not average cars.

3. The power output should be considered as Horsepower and Torque combined term

4. Regarding the specifications, considering all the conditions being same only power is calculated (Ideal stuff)
For example a Honda accord weighing around 1600kgs (including a bigger engine and all other stiffeners) power required will be at-least 700 bhp
Would have loved to clear it out but it seems the actual topic of discussion has nothing to do with what real world cars have to overcome, so i dont think its gonna apply here.

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Last edited by Shan2nu : 6th February 2010 at 01:25.
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Old 6th February 2010, 02:23   #32
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Wait a sec.

Does a 10 sec 1/4 mile produce 0.8Gs?

If you cover 400 meters in 10.9secs, your average speed is 132kmph. 132kmph = 120.29ft/s.

1/4 mile = 400meters = 1280ft.

(120.29 ft/s) sqrd / 1280 = 11.30ft/s sqrd.

1G = 32ft/s sqrd.

So 11.30 / 32 = 0.351Gs (Average).

Is this right?

But someone i have a feeling it should be somewhere around the 0.5 - 0.52G mark for a 10.9 sec 1/4 mile. Can anyone confirm this?

Shan2nu

Last edited by Shan2nu : 6th February 2010 at 02:42.
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Old 6th February 2010, 06:00   #33
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Originally Posted by Shan2nu View Post
Wait a sec.

Does a 10 sec 1/4 mile produce 0.8Gs?

If you cover 400 meters in 10.9secs, your average speed is 132kmph. 132kmph = 120.29ft/s.

1/4 mile = 400meters = 1280ft.

(120.29 ft/s) sqrd / 1280 = 11.30ft/s sqrd.

1G = 32ft/s sqrd.

So 11.30 / 32 = 0.351Gs (Average).

Is this right?

But someone i have a feeling it should be somewhere around the 0.5 - 0.52G mark for a 10.9 sec 1/4 mile. Can anyone confirm this?

Shan2nu
Metric system works fine for me.

Equation 2 of motion:-

s=ut+(at^2)/2 where s is the displacement, u is the initial velocity, t is the time, a is the acceleration.

400metres=(0 m/s)*11 s+[a(11 s)^2]/2

400 m = 121a/2

800/121=a=6.6 m/s^2 = 0.674 G.

Initially I had considered the time for a 10-second car to be 10 seconds which would have given an acceleration of 8m/s^2 or approximately 0.8G.

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Originally Posted by Shan2nu View Post

So you basically want a 1600kg car to accelerate at the rate of 8m/s^2 without "friction"?

Shan2nu
No, you missed the point. What I'm saying is that assuming friction wasn't present and assuming what I did in my first post was correct, a vehicle needs 690HP of power.

Now, suppose, in the real world, all the forces like friction, drag force etc. work against the vehicle, and it needs another 'x' amount of power to compensate for these, so that the power left to the vehicle after compensating for friction etc. is 690HP.

So total power required=690HP + x

Based on this, I was saying that power of more than 690HP is required in the real world. And in any case, a car wouldn't be able to move without friction.
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Old 6th February 2010, 06:11   #34
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And about the Ariel Atom, it's weight is merely 456kg, or 28.5% of the weight I've assumed in my calcs. So the power it requires to do the same would be 28.5% of 690HP in ideal conditions which is 196.65HP.

Seeing that it actually does it with 450HP, this means that a lot more power than an in ideal conditions is required to run a 10-second car.

With Veyron in ideal conditions [Mass=1888kg] power required is 814HP. The rest [1000HP-814HP=186HP] is used up in compensating for the drag, friction etc. Makes sense, I suppose.
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Old 6th February 2010, 11:24   #35
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No, you missed the point. What I'm saying is that assuming friction wasn't present and assuming what I did in my first post was correct, a vehicle needs 690HP of power.

Now, suppose, in the real world, all the forces like friction, drag force etc. work against the vehicle, and it needs another 'x' amount of power to compensate for these, so that the power left to the vehicle after compensating for friction etc. is 690HP.

So total power required=690HP + x

Based on this, I was saying that power of more than 690HP is required in the real world. And in any case, a car wouldn't be able to move without friction.
But even if we consider your calculations to be correct, is there anyway you could actually make this happen in a lab, with a 18:1 scale model weighing 88.88kgs, producing 38bhp, over a distance of 22.22mtrs?

You can remove aero resistance with vacuum, but friction im not sure can be taken out of the equation coz you need friction to even get the car moving.

Maybe you could use mag-lev technology to reduce friction.

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And about the Ariel Atom, it's weight is merely 456kg, or 28.5% of the weight I've assumed in my calcs. So the power it requires to do the same would be 28.5% of 690HP in ideal conditions which is 196.65HP.

Seeing that it actually does it with 450HP, this means that a lot more power than an in ideal conditions is required to run a 10-second car.

With Veyron in ideal conditions [Mass=1888kg] power required is 814HP. The rest [1000HP-814HP=186HP] is used up in compensating for the drag, friction etc. Makes sense, I suppose.
Ok im just wondering, without drag and friction, does the weight of the vehicle even matter? Bcoz the main reason why a heavier load bcoms diff to pull is bcoz its weight creates a higher amount of rolling resistance/friction.

Its like they say, if you drop a piece paper and a piece of iron in a vacuum, they will fall at the rate of 9.8m/s sqrd, irrespective of their mass, size or shape.

So your theory seems a bit contradictory wrt friction. You can't have friction only where you want it (which is to propel the car over 400mtrs in 10.9secs) and ignore its disadvantages (which is the drag caused by the very laws of friction letting your car move forward).

Shan2nu

Last edited by Shan2nu : 6th February 2010 at 11:37.
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Old 6th February 2010, 11:47   #36
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Quote:
Originally Posted by anku94 View Post



You're right. I was talking about a model engine or whatever. And you were again right. I'm heavily inspired by the Fast and Furious series
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Originally Posted by McLaren Rulez View Post
The OP has assumed his engine to have a constant power at all rpm. So why bring torque or engine size into it? Constant and known power implies that the torque is also constant and known. If he's talking about a real engine, then we can talk about torque. But he's using a model engine with constant power.

I think the OP was surprised at the fact that a ten second car needs so much power. But that is the truth. The Fast and Furious movies make ten second cars appear a bit too commonplace...

not to trivialize your discussion or discourage you, but that's exactly the difference in theory and real life. That's why we go to our physics labs hoping to measure real quantities with formulae in hand and nothing adds up.

you can't create a model with model engine and no losses, and then be amused by the fact that the real world cars do not follow your model .

All your theory will tell you that you need "at least" that much of average power. How much to add for losses, what BHP to consider for the RPM range it is going to operate etc will only be discovered in an actual experiment. In all probability your xxx BHP car may never hit that BHP by the time it finishes your short test run.
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Old 6th February 2010, 12:16   #37
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Quote:
Originally Posted by Shan2nu View Post
Wait a sec.

Does a 10 sec 1/4 mile produce 0.8Gs?

If you cover 400 meters in 10.9secs, your average speed is 132kmph. 132kmph = 120.29ft/s.

1/4 mile = 400meters = 1280ft.

(120.29 ft/s) sqrd / 1280 = 11.30ft/s sqrd.

1G = 32ft/s sqrd.

So 11.30 / 32 = 0.351Gs (Average).

Is this right?

But someone i have a feeling it should be somewhere around the 0.5 - 0.52G mark for a 10.9 sec 1/4 mile. Can anyone confirm this?

Shan2nu
How can you say that the car will accelerate at constant values ie,. average 120.29 ft/sec

Acceleration of the car will be non linear and very much depends on the gear ratios, transmission etc
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Old 6th February 2010, 12:24   #38
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How can you say that the car will accelerate at constant values ie,. average 120.29 ft/sec

Acceleration of the car will be non linear and very much depends on the gear ratios, transmission etc
Snap out of the real world dude. We're no longer there. LOL

The acc of the therotical car is supposed to be linear.

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Old 6th February 2010, 12:42   #39
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Originally Posted by Shan2nu View Post

Ok im just wondering, without drag and friction, does the weight of the vehicle even matter? Bcoz the main reason why a heavier load bcoms diff to pull is bcoz its weight creates a higher amount of rolling resistance/friction.

Its like they say, if you drop a piece paper and a piece of iron in a vacuum, they will fall at the rate of 9.8m/s sqrd, irrespective of their mass, size or shape.
Of course the weight matters! Or to be more specific, the mass matters. Because acceleration will vary with mass once your force is constant. Its exactly what Newton's second law talks about.

When you talk about falling objects, the force acting on a body falling towards earth depends on its mass. Heavier objects are pulled with a greater force by the earth. Our engine's force is a constant so its a different case from a falling body

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Originally Posted by Shan2nu View Post
So your theory seems a bit contradictory wrt friction. You can't have friction only where you want it (which is to propel the car over 400mtrs in 10.9secs) and ignore its disadvantages (which is the drag caused by the very laws of friction letting your car move forward).
Shan2nu
I think you're saying there has to be rolling friction to enable the wheels to get traction. In his case, the model doesn't have wheels that roll if you get what I mean. It simply says you have a mass m and you exert a force F and see what F must be for a given m to get the necessary acceleration. It will be easy enough to make a model that includes rolling friction. I can do it if anyone wants.

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Originally Posted by vivekiny2k View Post
you can't create a model with model engine and no losses, and then be amused by the fact that the real world cars do not follow your model .

All your theory will tell you that you need "at least" that much of average power. How much to add for losses, what BHP to consider for the RPM range it is going to operate etc will only be discovered in an actual experiment. In all probability your xxx BHP car may never hit that BHP by the time it finishes your short test run.
Yeah but to create a realistic model is going to be incredibly hard. The sheer number of variables and energy losses will baffle anyone working with a simple calculator.
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Old 6th February 2010, 12:42   #40
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Sure enough, no number of equations can describe the real world. Yet, the OP needs to add a few degrees of complication to his model in order to arrive at a slightly more credible answer.

Firstly, assuming no friction is too much of a simplification. So please factor in:
1) rolling resistance
2) aerodynamic drag (assume a standard frontal area and a cd of ~ 0.3)

Then, you need to consider that no engine makes the same power throughout its rev-range. So take a look at the power/torque curves of several cars, and then create a model of a generic curve. Base your acceleration calculations on this curve. You will need to step beyond linear equations of motion and use calculus. But calculus is fun!

If you can do all that, the credibility of your calculation will increase dramatically.
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Old 6th February 2010, 13:07   #41
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Of course the weight matters! Or to be more specific, the mass matters. Because acceleration will vary with mass once your force is constant. Its exactly what Newton's second law talks about.

When you talk about falling objects, the force acting on a body falling towards earth depends on its mass. Heavier objects are pulled with a greater force by the earth. Our engine's force is a constant so its a different case from a falling body
In the absense of any resistance, i dont think it does.

If we were to imagine the gravitational pull of earth was equivalent to a 686bhp engine and has a gravitation rate of 6.6m/s/s.

In such a situation, the mass of the object should not affect the rate of acceleration.

Bcoz like the earths gravitational pull, the pulling power this 686bhp motor remains contant (since you've mentioned that it does 686hp at any given rpm). But in the real world factors like aerodynamic and frictional resistance affect its rate of acc (which in this experiment are non existant). So just like a free falling object in vacuum, there is only 1 force acting on the car and that is the pulling power of the engine. Weight makes no sense without the presence of friction.

The acceleration of the object equals the gravitational acceleration. The mass, size, and shape of the object are not a factor in describing the motion of the object. So all objects, regardless of size or shape or weight, free fall with the same acceleration. In a vacuum, a beach ball falls at the same rate as an airliner. Knowing the acceleration, we can determine the velocity and location of any free falling object at any time.

Source - Free Falling Object

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Last edited by Shan2nu : 6th February 2010 at 13:16.
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Old 6th February 2010, 13:14   #42
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If we were to imagine the gravitational pull of earth was equivalent to a 686bhp engine and has a gravitation rate of 6.6m/s/s.
No, this is an incorrect assumption. The earth exerts different forces on different masses. So the gravitational pull cannot be a constant. Newton's law of gravitation, remember? The acceleration is a constant but the force is not.

Last edited by McLaren Rulez : 6th February 2010 at 13:16.
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Old 6th February 2010, 13:34   #43
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No, this is an incorrect assumption. The earth exerts different forces on different masses. So the gravitational pull cannot be a constant. Newton's law of gravitation, remember? The acceleration is a constant but the force is not.
But if there is no friction between the car and the ground. Whats stopping the engine from pulling the car at its max rate of acc?

If you take a fixed amount of force acting on an object weighing 1000kgs placed on tarmac, the same amount of force acting on the same object placed on ice and again the the same force acting on the same object levitating.

The force acting on it is the same, the weight of the object is the same, but will the rate of acceleration be the same as well?

If you compare a 1000kg object on tarmac and a 1500kg object on ice and let the same amount force act on them. You could see both objects accelerating at similar rates even though their mass is different.

We've all played "Carrom" at some point in our lives. Why do you need to apply more force to push the striker a given distance in a given time in the absence of powder? And how come the moment powder is applied to the board, that same striker (which has not changed its mass/weight) accelerates at the same rate as before, with just a fraction of the force required earlier?

So the force required to pull/push an object with a fixed mass over a specified distance, reduces with reduction in friction, but what if there was no friction at all? How would a 1000kg object offer more resistance than a 1kg object when a horizontal force is applied to it?


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Last edited by Shan2nu : 6th February 2010 at 13:46.
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Old 6th February 2010, 13:46   #44
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Ok let me elaborate.

First, I hope you realized why the falling body analogy is wrong. If not, tell me and I'll explain more clearly. From now on, I will assume everwhere that there is no resistance due to friction unless specifically stated otherwise.

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But if there is no friction between the car and the ground. Whats stopping the engine from pulling the car at its max rate of acc?
The precise term in inertia. You need to apply a force to move an object and its acceleration will depend on its mass. A heavy object will accelerate slowly compared to a light object given the same force.

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Originally Posted by Shan2nu View Post
If you take a fixed amount of force acting on an object weighing 1000kgs placed on tarmac, the same amount of force acting on the same object placed on ice and again the the same force acting on the same object levitating.

The force acting on it is the same, the weight of the object is the same, but will the rate of acceleration be the same as well?
Since tarmac and ice don't offer any friction, the acceleration will be identical. Levitation introduces gravity, an extra force, so that will be different.

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Originally Posted by Shan2nu View Post
If you compare a 1000kg object on tarmac and a 1500kg object on ice and let the same amount force act on them. You could see both objects accelerating at similar rates even though their mass is different.
No this would not happen. Since there is no friction, tarmac and ice are indistinguishable. So the 1000 kg object would accelerate faster.

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Originally Posted by Shan2nu View Post
So the force required reduces with reduction in friction, but what if there was no friction at all? How would a 1000kg object offer more resistance than a 1kg object when a horizontal force is applied to it?

Shan2nu
As mentioned earlier, the 1000kg object has more inertia than the 1kg object. That is the basis of Newton's second law. F=ma because a higher m for a given F will result in a lower a.

End of high school physics class
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Old 6th February 2010, 13:51   #45
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You guys are all digressing too much into automobile jargon here. The question was a simple one and he has answered it himself.
Yes anku94 with the assumptions you have made it requires that amount of power to go that distance in that time.

Guys keep one thing in mind don't ignore physics when in real world!
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