Hate to post too many threads but this video is amazing

[dacoyote]

8.83 seconds!!!

Unbelievable. Can someone do a little math and tell us how many G of acceleration that is (1G = 32ft/sec/sec or 9.8m/sec/sec)?

Looks like he has a 'chute, though we never see it deploy.

That is , fast. That car is probably an E34 is outward appearance only. Sick.

Assuming it went to 253Kph in 8.83 seconds....

Then assuming it speed up at 28.65 by kphps

Thus giving us 7.9583 m / s

Thus giving us .812 Gs

[Jay 535i]

Assuming it went to 253Kph in 8.83 seconds....

Then assuming it speed up at 28.65 by kphps

Thus giving us 7.9583 m / s

Thus giving us .812 Gs

Gee, I'd have expected at least 1G.

Thanks, Mathman.

[dacoyote]

Gee, I'd have expected at least 1G.

Thanks, Mathman.

So would I... I checked my math twice... and even did it a different way.... both answers where within about a tenth of a G.. (rounding errors)

[Jay 535i]

So would I... I checked my math twice... and even did it a different way.... both answers where within about a tenth of a G.. (rounding errors)

Oh, I see your error. You divided the trap speed by the ET, giving you the average acceleration: 28.65 kph/s

The acceleration off the line would've been much greater than the average over the whole 1/4 mile.

Now that I think of it, I think there's insufficient information to derrive the max G, because aerodynamics would be a factor.

[DanH]

Sorry to break it to you, but that is not how you calculate the acceleration, better brush off the dust from the ol' physics book:

The formula for average acceleration (a) is:

a = (v*v - u*u)/ (2 *d)

Assume:
v= final velocity = 253kph = 70.27 m/s
u= initial velocity = 0
d= is distance travelled = 1/4 mile = 402.33 m

Then:
a = (70.27 * 70.27 - 0) / (2 * 402.33) = 6.14 ((m/s)/s)

Divide by 9.8 ((m/s)/s) to get G force equivalent.

Which is about 0.63 Gs. Not quite so much as many of you expected.

Its been about 7 years since I had to do any physics problems, but I knew something was dacoyotes numbers, so I looked up the formula.

Assuming it went to 253Kph in 8.83 seconds....

Then assuming it speed up at 28.65 by kphps

Thus giving us 7.9583 m / s

Thus giving us .812 Gs

[Jay 535i]

Sorry to break it to you, but that is not how you calculate the acceleration, better brush off the dust from the ol' physics book:

The formula for average acceleration (a) is:

a = (v*v - u*u)/ (2 *d)

Assume:
v= final velocity = 253kph = 70.27 m/s
u= initial velocity = 0
d= is distance travelled = 1/4 mile = 402.33 m

Then:
a = (70.27 * 70.27 - 0) / (2 * 402.33) = 6.14 ((m/s)/s)

Divide by 9.8 ((m/s)/s) to get G force equivalent.

Which is about 0.63 Gs. Not quite so much as many of you expected.

Its been about 7 years since I had to do any physics problems, but I knew something was dacoyotes numbers, so I looked up the formula.

We're not interested in the average -- just the max, which you can't know without factoring in the car's Cd.

[dacoyote]

Oh, I see your error. You divided the trap speed by the ET, giving you the average acceleration: 28.65 kph/s

The acceleration off the line would've been much greater than the average over the whole 1/4 mile.

Now that I think of it, I think there's insufficient information to derrive the max G, because aerodynamics would be a factor.

yeah... thats part of it.... it's a average.....

The video pan's so I cannot tell from that the curve....

.8G's the entire way isn't anything to poopoo.....

[DanH]

I had another reply but it disappeared, i prob hit preview post instead of submit reply...

You are correct that we need more info to calculate max accel. I was just pointing out the avg. accel. calculation error.

What you need is data of instantaneous speed vs. time and then take the derivative (now you need that calculus book) of that to determine the instantaneous acceleration vs. time. Find the max value on that chart and you'll have your answer.

We're not interested in the average -- just the max, which you can't know without factoring in the car's Cd.

[Jay 535i]

I had another reply but it disappeared, i prob hit preview post instead of submit reply...

You are correct that we need more info to calculate max accel. I was just pointing out the avg. accel. calculation error.

What you need is data of instantaneous speed vs. time and then take the derivative (now you need that calculus book) of that to determine the instantaneous acceleration vs. time. Find the max value on that chart and you'll have your answer.

I'll settle for "it's really fast."

[dacoyote]

.....
a = (v*v - u*u)/ (2 *d)
.....

same thing I used... not sure why my numbers where so far off....

I did it on paper after a couple drinks :-)