Calculating estimated HP from velocity, auto weight, and constant acceleration I am working on a simulation program that runs theoretical performances of different cars and was wondering if there is a way to estimate the HP at any running RPM?
the problem is (and this may be me being naive) but I run the code and in first gear the HP at 6200 RPM is approx 48 on a car rated at 106 HP. Now I would guess the HP at the wheels would be slightly lower then the engine rated HP, but not by over half.
The data I have is:
mass: 1143 kg;
HP: 106 @ 6200 RPM;
First gear ratio:  3.25
Final Gear: 4.25
wheel and tire circumference: 1.8743 m
I set the constant acceleration to the velocity at 1000 RPM figuring that at 0 RPM we have 0 acceleration and at 1000 RPM we have x m/s so that average of acceleration is x m/s/s.  Again I almost feel I am wrong with this assumption so correct me if so. 
My process was to convert RPM at the engine to RPM at the wheels (assuming no wheel slip and 100% efficiency of drive line). then converted RPM @wheels to m/s
then with constant acceleration assumption, the force to move the car is m * a
and then converted the Newtons to power in watts then watts to HP (1 W = 745.7 HP).  
 A: It's a little bit hard to follow your reasoning. Let me try to give the method I would use - simple balance of energy.
If the car reaches a velocity $v$ after time $t$, the power of the engine will have been used mostly for five components:


*

*Kinetic energy of car: $\frac12 m v^2$

*Rotational kinetic energy of tires: $4 \times \frac12 I_t \omega_t^2$

*Rotational kinetic energy of engine: $\frac12 I_e \omega_e^2$

*Rolling friction: $mg\mu d$

*Air drag - becomes significant as speed increases.


The rotational component of the tires is not insignificant: a uniform cylinder of mass $m$ and radius $r$, rolling with velocity $v$ has angular velocity $\omega = \frac{v}{r}$ and rotational kinetic energy 
$$\begin{align}KE &= \frac12 I \omega^2\\
&=\frac12 (\frac12 m r^2 ) \left(\frac{v}{r}\right)^2\\
&=\frac14 m v^2\end{align}$$
In other words, the total kinetic energy of the tires (linear plus rotational) is 50% greater than just the linear component (for a uniform tire). In other words, it's like carrying the mass of two additional tires in the car. 
Similarly, the engine, running at 6000 rpm = 100 revs/sec, is storing a significant amount of energy as well - if you think about it, when you rev an engine it takes a bit of time to get up to speed, which tells you that maybe half a second's worth of engine power is spent just bringing the engine up to speed.
The other components you would have to estimate from the numbers you have - it's a bit hard to understand the distance your car moved, or how long it was moving for.
The air drag is a function of the square of the velocity, so it will be most significant near the top speed of the car. You can write the instantaneous force of drag as
$$F=\frac12 c_d \rho A v^2$$
where $c_d$ is typically around 0.3 and $C_d A$ around 0.55 m2. With $\rho = 1.3 kg/m^3$, the force is approximately
$$F = 0.27\times \left(\frac{v}{3.6}\right)^2$$
with $v$ in km/h - meaning that at 100 km/h, the force is for a streamlined car is a little over 200 N, and the power needed to overcome that drag ($P = Fv$) is around 6 kW, or 7.5 hp.
For rolling resistance, car tires have a coefficient around 0.03, so the force (for your 1143 kg car) is
$$F_{rolling}=mgc=336 N$$
and the power used is
$$P = Fv = 9.3 kW ~ 12 hp$$
For a speed of 100 km/h, and a tire mass of 30 kg / tire, I compute kinetic energy of 
$$KE = \frac12(m_{car} + 2m_{tire})v^2 = 464 kJ$$
Forgetting for a moment the power curve of the engine, if the car accelerates to 100 km/h in 10 seconds, the distance covered will be
$$d = \frac12 a t^2 ~ 138 m$$
At this point, the energy spent rolling was
$$E_{roll} = 336\times 138 = 46.7 kJ$$
Integrating the air drag over 10 seconds:
$$E_{drag} = \int F(v) v dt \\
= \int 0.27 \left(v(t)\right)^3 dt \\
= \int 0.27 \left(at\right)^3 dt \\
= \frac{0.27\times 2.8^3}{4}t^4\\
~14.8 kJ$$
The air drag will be less (since the drag increases with square of velocity, and is less than rolling friction even at 100 km/h), so the major component will be the kinetic energy of the car.
Just from the above, the car that takes 10 seconds to accelerate has spent 464+47+15 = 526 kJ, or 52.6 kW - meaning an average power output of around 71 hp.
As you can see, the time to accelerate really matters if you want to use acceleration to get the power. It scales roughly inversely - but there's quite a bit to making this even vaguely accurate. And the "power output" of an engine will be a strong function of rpm, so while it may peak at 6800 rpm, that's not where it spent a lot of the accelerating time...
I hope this gives you food for thought - good simulation takes time.
