Math boffins, pit time question:

Not average speed - a gap in average speed. If the average speed (integral( v*dt) / T with the lap time T) of car A is 150km/h and the one of car B is 160km/h, their temporal distance increases.

Edit: Or is it possible that I'm nearing the limits of my language skills in terms of what 'gap' means?

 

The initial poster did not refer to a gap in average speed, meaning a difference between the average speed of one car relative to another. The question assumed identical average speeds, therefore a constant time gap between the two cars at a particular point on the circuit where that gap was measured, and how that gap would change when another event occurred, a difference in pit stop times for both cars.
Clearly, unless the cars were circulating at a constant speed, both the time and distance gaps would alter at different points of the circuit as the cars slowed and then accelerated to negotiate corners.

 

Yes, but I was quoting David, who was talking about... *drum roll*... a constant gap in average speed!

 

My English is worse than yours.
By saying that the gap in average speed remained constant I was meaning to say that the average speeds (lap times) were the same.
The point being that the physical gap between cars would vary as one sped up or slowed down for corners even if the total elapsed lap times were identical.

 

By our previous discussion I already supposed you wanted to say that, but the academic in me just couldn't let an imprecise statement left uncommented *sternly looking over the rim of my glasses*

 

Show some respect for your elders.......

 

It's been a while since I've been taken to the wood shed.

 

I thought you'd go there regularly with Steve... (cue Jane Birkin & Serge Gainsbourg!)

 

I think that you're unfamiliar with the idiom.

 

http://www.usingenglish.com/reference/idioms/take+someone+to+the+woodshed.html
Don't worry Prof Florian math boffin I'll teach you..

And I'm not talking MAXXX Mosley style..

 

Your chance to return the evil stare now! In a duet with Steve!


Back to topic, I find the idea of the article that Ian linked - the calculation of an optimal pit stop strategy - quite interesting, although I wouldn't do a Fuzzy approach. Is there any interest in discussing the topic in a more detailed matter? If yes, I'd dig out some lap time data tomorrow and sit down and think of how we could model it...

 

Interesting turn that a "scientific" thread has taken.


One of the first questions I have when watching a GP is what the pit delta is.
That races are won and lost there is an old story but holds true.

 

Absolutely. And what I find rather problematic too is the importance of the order the pits are arranged in when it comes to the (necessary!) penalties for "unsafe release".

 

Unsafe release?
Is that code?


This is a variation on the original question.
Pit spaces are allocated by championship rankings.
One end considered to hold an advantage over the other.
But the total distance a pit stop covers is identical
given the pit speed limit average speed (time in motion) should be the same as well.
So what is the placement advantage?

 

The advantage - which I perceive, maybe it doesn't exist! - is when the team closer to the pit entrance is slightly faster. Not fast enough for their driver to 'regularly' overtake the other one, but enough for him to be already in the lane and thus forcing the others to hold their guy back if they don't want to risk the unsafe release.

Edit: And I'm scared by the potential multiple meanings of "unsafe release" which you imply

 

As the man taking the Rorschact test said. "but Doc, they're your dirty pictures!".

 

Simplify . . .

1 car, not 2 cars.

1 car, 2 pit stops.
________________

Car pits for 3 seconds.

Lap speed, momentum, inertia, gravity is maintained, etc.

Car pits for 3.5 seconds.

Lap speed, momentum, inertia, gravity is maintained equal to previous stop, etc.

.5 second difference.
_________

Same thing if there were 2 cars.

.5 second difference.

 

Must be the variable gravity that threw us off

 

If there is a treadmill in the pitlane can the cars still take off?

>8^)
ER

 

Speaking of taking off.
Given their level of grip why not run a race on a suspended track shaped as a möbius strip?

 

A duet, or duel?

Sign me up! - It's a fascinating area for the geeks I guess! FWIW, my handy-dandy, super expensive timing app has all the data downloaded from the race. I'd need a lap chart to easily zoom in onto in laps etc, but reckon we've got a good starting place.

I believe you're correct - Cars 1&2 always get the first box and it goes down from there. If I'm, say, #1 and you're right up my chuff on entry I stop before you, and hence get going again before you and can thence block your exit. [All else being equal of course.]

Cheers,
Ian

 

I think it depends on the track regarding the most advantageous spot. I believe the current championship team get to choose if they the want the first or last pit spot and everyone else goes from there in order of LY points

 

For reasons I can't now remember, I do believe you're correct - IIRC, at least Silverstone (& Indy?) were set up that way.

So, the question becomes, what's the advantage of doing it 'the other way'? If I'm at the far end, *and* I spend less time in the air than you, I can get out ahead. But, empirically, I still think I want to stop first and hence start first...... Does that change depending on the delta? ( I can't think why, but am going all Yoda on it now......)

..... Wheres Florian

Cheers,
Ian

 

This thread makes my head hurt

 

Yup, giving us the so-called (by SPEED anyway) concertina effect, where cars get spatially closer in braking, but not temporally (ie, time-wise, not a typo for temporarily) closer.

This is a good time to mention the illusion of a car "pulling away" from another out of a corner. At 1m/s, a gap of 1s = 1m. At 20m/s, a gap of 1s = 20m. If 2 equally driven cars accelerate out of a corner at these speeds, the lead car appears to be pulling away but even though he is strictly maintaining a 1s gap, he extends the linear distance by 20 times! It's the reverse of the concertina effect and often gives a false impression to the viewer.

As to the 0.5s thing, it's important to remember it's not just the time "in the box", it's the entire time in the pit lane that matters.