Accelerating a moving object with force

[magnethead494]

Spreadsheet of data: http://i1105.photobucket.com/albums/h355/magnethead494/Screenshot2012-05-14at103445PM.png

Example dyno graph: http://i1105.photobucket.com/albums/h355/magnethead494/turbo_busa_dyno_hahn_stg1.jpg

I'm familiar with F = MA/G => F*G/M = A for english units (G = 32.17).

This works for accelerating from a standstill (IE, approx'ing to top of first gear),

160 ft-lb * 1.596 primary * 2.615 first gear * 4.0625 chain drive = 2,712 ft-lb at the axle
2,712 * 12 inches / 13.25 inch rolling radius = 2,456 pound force
2,456 pound-force / 700 pound-weight = 3.50 G-forces = 112.87 feet per second squared average acceleration

I know that seems like a high acceleration, but for a 300HP 700 pound machine, that's a 2.3 pound-per-HP ratio.

And the formula

(V_f)² = (V₀)² + 2ad

for subsequent gears... But I don't know x.

Applying that formula to previous calculation,

(75.68 feet/sec)² = (0MPH)² + 2 * 112.87 fps² * d

5,727.4624 feet² per second² = 225.74 fps² * d

5,727.4624 feet² per second² / 225.74 feet per second² = 25.372 feet to the top of first gear (AKA 25.372 feet to 51.6 miles per hour, at average 3.5G's)But how can I calculate the acceleration for subsequent gears?
I'm going to state my 3-part hypothesis..am I right or wrong?

Part 1: Using F=Ma to get acceleration per gear

Expanded:

top of gear:
primary * chain = 6.48375
rolling radius multiplier = 0.88888
Engine Torque at 10,500 RPM: 160 ft-lb
Composite value = 160 * 6.48375 * 0.8888888 = 922.13333
1st: 922.1333 * 2.615 / 700 * 32.17 = 110.82 feet per second squared (3.44G)
2nd: 922.1333 * 1.937 / 700 * 32.17 = 82.08 feet per second squared (2.55G)
3rd: 922.1333 * 1.526 / 700 * 32.17 = 64.67 feet per second squared (2.01G)
4th: 922.1333 * 1.285 / 700 * 32.17 = 54.45 feet per second squared (1.69G)
5th: 922.1333 * 1.136 / 700 * 32.17 = 48.14 feet per second squared (1.50G)
6th: 922.1333 * 1.043 / 700 * 32.17 = 44.20 feet per second squared (1.37G)

Part 2: Getting the distance per gear

1st: 75.68² = 0² + 2 * 110.82 * d => 5,727.46/221.64 = 25.84ft
2nd: 102.18² = 75.68² + 2 * 82.08 * d => 10,440.75 - 5,727.46 = 164.16 * d => 4,713.29 / 165.16 = 28.53ft
3rd: 129.71² = 102.18² + 2 * 64.67 * d => 16,824.68 - 10,440.75 = 129.34 * d => 6,383.93 / 129.34 = 49.35ft
4th: 154.03² = 129.71² + 2 * 54.45 * d => 23,725.24 - 16,824.68 = 108.90 * d => 6,900.56 / 108.90 = 63.36ft
5th: 174.24² = 154.03² + 2 * 48.14 * d => 30,359.58 - 23,725.24 = 96.28 * d => 6,634.04 / 96.28 = 68.90ft
6th: 189.77 = 174.24² + 2 * 44.20 * d => 36,012.65 - 30,359.58 = 88.40 * d => 5,653.07 / 88.40 = 63.95ft

25.84 + 28.5 + 49.35 + 63.36 + 68.90 + 63.95 = 299.9 feet (should equal something near 660ft)

Something is clearly wrong with the calculation...but what is it? The G forces seem realistic, the MPH calculations (spreadsheet and google to FPS calcs) seem correct, what else could be wrong?

Part 3: getting the time per gear

d = t/2 * (V_f+V₀)

1st: 75.68² = 0² + (2 * 110.82 * t/2 * (75.68 + 0)) => 5,727.46 = 8,386.86 * t => 0.683 seconds
2nd: 102.18² = 75.68² + (2 * 82.08 * t/2 * (102.18 + 75.68)) => 10,440.75 = 5,727.46 + 14,598.75t => 0.323 seconds
3rd: 129.71² = 102.18² + (2 * 64.67 * t/2 * (129.71 + 102.18)) => 16,824.68 = 10,440.75 + 14,996.33t => 0.426 seconds
4th: 154.03² = 129.71² + (2 * 54.45 * t/2 * (154.03 + 129.71)) => 23,725.24 = 16,824.68 + 15,449.64t => 0.447 seconds
5th: 174.24² = 154.03² + (2 * 48.14 * t/2 * (174.24 + 154.03)) => 30,359.58 = 23,725.24 + 15,802.92t => 0.420 seconds
6th: 189.77 = 174.24² + (2 * 44.20 * t/2 * (189.77 + 174.24)) => 36,012.65 = 30,359.58 + 16,089.24t => 0.351 seconds

0.683 + 0.323 + 0.426 + 0.447 + 0.420 + 0.351 = 2.65 seconds (should equal something near 4.95 seconds)

Again, something is clearly wrong. I have a feeling it's in the main equation...(V_f)² = (V₀)² + 2ad

A = G * F/m

d = t/2 * (V_f+V₀)

If someone could point my mistakes, it would be appreciated.

 

[haruspex]

You've neglected air resistance. That would be pretty significant long before you reach top speed.

 

[magnethead494]

You've neglected air resistance. That would be pretty significant long before you reach top speed.

One would think so, but it's surprisingly not as big an issue as one would think. The body panels and ground effects are done up to where, once you have the power on tap, air resistance is a very small loss, maybe 4% on MPH.

Either way, that's nothing to do with the question I asked. I asked where my calculations were off by 200% or so.

 

[haruspex]

Then you're assuming constant power (300HP) and no drag.
In that case we can skip the details and go straight to the energy.
A 700 pound (318kg) mass at 189.77 fps (57.86 mps, 129Mph) has 532.5kJ.
A power of 300HP (225000 W, or thereabouts) delivers that in 2.37 seconds.
If that is not reflected in reality then you must be ignoring a significant loss somewhere.

 

[PJC01]

I can offer some insights and clarification to your calculations and hypotheses.

Firstly, your calculation for acceleration per gear using the formula F=Ma is correct. However, it is important to note that this formula assumes a constant force acting on the object. In reality, the force applied to a moving object may vary, especially in the case of a vehicle accelerating through different gears. This could affect the accuracy of your calculations.

Secondly, your calculation for distance per gear is not entirely correct. The formula you used, (Vf)2 = (V0)2 + 2ad, assumes a constant acceleration. In the case of a vehicle accelerating through different gears, the acceleration is not constant and therefore this formula cannot be used. Instead, you can use the formula d = (Vf+V0)/2 * t, where t is the time it takes to reach the final velocity (Vf) from the initial velocity (V0). This will give you a more accurate calculation for distance traveled.

Lastly, your calculation for time per gear is also incorrect. The formula you used, d = t/2 * (Vf+V0), is the rearranged version of the formula d = (Vf+V0)/2 * t, which I mentioned above. However, you also need to take into account the acceleration of the vehicle in this formula. The correct formula would be t = (Vf-V0)/a, where a is the average acceleration of the vehicle in that gear.

In summary, your main mistake seems to be assuming constant acceleration in your calculations. In reality, the acceleration of a vehicle will vary as it accelerates through different gears, so it is important to take that into account in your calculations. Additionally, the force applied to a moving object may also vary, which can affect the accuracy of your calculations.

I hope this helps clarify some of your concerns and aids in your understanding of the physics behind accelerating a moving object with force. Keep up the scientific thinking and good luck with your research!