oval speed calculation

[John Finch]

Here is one I just can't find the logic for. I hold the e hydro record at 17 seconds for two laps, so I was there when it happened. I understand and saw how it was done. LOL. The boat was running between 98 and 100 mph on the radar gun. The boat ran 8.5 seconds per lap. Here is what I don't understand.... The record certificate says the boat traveled at 70 mph average speed. So here is the math.... It takes 2.2 seconds to travel 330 feet at 100 mph. That is the length of the straightaway from buoy six to buoy one. It Doesn't matter how close to the buoys........330 feet is 330 feet. If the boat was running 100 mph and it traveled the front and back straightaway at 100 mph that's 4.4 seconds of the 8.5 total for one lap. That only leaves 4 seconds to go around the corners, so that equates to 2 seconds per corner. Seconds vs mph figures ok at 100 mph as I saw it happen.

Now, lets use the IMPBA rule book chart to see how the theoretical speed is calculated.....

According to the IMPBA chart in the rule book it takes 3.2 seconds to travel the 330 foot straightaway at 70 mph. That's 6.4 seconds of the 8.5 total, so that means the boat would have to get through each of the two corners at just a tad over one second. I don't care how close you hang on the buoys......I don't see a one second trip through the corners at 70 mph. It just does not figure to me.

 

[Scott Norris]

I take it you are reading something from the IMPBA rule book John. What page are you looking at??

 

[John Finch]

The calculations are at the end of the book after record trial info. The book I am looking at was revised 1997 page c12 and c14. The facts just don't match the theoretical stuff in the book. It is embarrassing when you have a guest come to your house, you tell them a boat does 100 mph and he reads the 70 mph speed on the certificate and thinks you are telling a big fish story. This happened to me this week during a party my wife had for her people at work. One of the guests wanted to see the boats and noticed the certificate for the E record only read 70 mph after I told him the boat was doing 100.

 

[Scott Norris]

If you cover 880 feet in 8.5 seconds, that's 70.579936 MPH. I'm impressed, if your friends aren't maybe they should come out and see your boats run!!

A 3 foot long boat going 100 down the straights is something to behold!!!

 

[Don Ferrette]

John it's from the same mathbook that some out there are using to try to convince everyone the new voting proposal is better for all. :lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol:

Seriously, it has never looked right to me either. Our team oval record in D hydro says 69.436 mph when we ran 109 with same boat 2 weeks earlier at SAW. The F hydro oval record you & Kentley set says you went 72.381 when we had you on radar hitting the corners at well over 100. Even though the math may work it out I'd just assume the speed wasn't even on the record as it makes people think they can do this with their heat race boats. :blink:

 

[piper_chuck]

They're obviously losing alot of speed in the turns, although I'm sure it doesn't feel that way when you're the one holding the transmitter.

 

[Scott Norris]

Going into a corner at 100 plus would scare the $#&( outa me.

 

[John Finch]

If you cover 880 feet in 8.5 seconds, that's 70.579936 MPH. I'm impressed, if your friends aren't maybe they should come out and see your boats run!!

A 3 foot long boat going 100 down the straights is something to behold!!!

Scott,

So how many feet would the boat be traveling at 100 mph to do a 8.5 sec lap? Must be way off the buoys, or is just a little off the buoys a LOT MORE FEET? Pull out that calculator. During the F record the boat was clocked at 103 at the end of the straightaway and 98 coming out of the corners, so figure an average of 100 mph and figure the feet traveled. Please.

 

[Don Ferrette]

They're obviously losing alot of speed in the turns, although I'm sure it doesn't feel that way when you're the one holding the transmitter.

That's the odd part, 5 to 6mph with the hydros & this was on radar. That's why it just doesn't look right. :blink:

 

[Scott Norris]

1246.6666 feet at 100mph over 8.5 seconds

 

[Don Ferrette]

If you cover 880 feet in 8.5 seconds, that's 70.579936 MPH. I'm impressed, if your friends aren't maybe they should come out and see your boats run!!

A 3 foot long boat going 100 down the straights is something to behold!!!

Scott,

So how many feet would the boat be traveling at 100 mph to do a 8.5 sec lap? Must be way off the buoys, or is just a little off the buoys a LOT MORE FEET? Pull out that calculator. During the F record the boat was clocked at 103 at the end of the straightaway and 98 coming out of the corners, so figure an average of 100 mph and figure the feet traveled. Please.

And I can tell you from first hand experience, John was shaving a coat of paint off he ran so close to the buoys. That's why we did the D hydro oval record as a team, when Finch hits his groove at oval record trials there ain't no one better. B)

 

[Scott Norris]

I agree it doesn't seem right.

 

[DON MAHER]

Hey John,

This is the math I was given to figure MPH for a 1 mi. 2-lap corase. 29.33 divide by time in sec. times 20 divide by 0.488 = MPH. I know it dosent sound right, but it really is. Doing this math a 2-lap time of 17.00 sec. = 70.708 MPH. I think what you may not have accounted for is decceleration time thru the turns & the acceleration time back to 100 MPH.

Don

 

[Scott Norris]

That's almost a half a lap difference.

 

[John Finch]

1246.6666 feet at 100mph over 8.5 seconds

ok, so the boat traveled 366 feet further than it had to. That's more than a full straightaway!

How about this. Can you figure out how far the boat would have to travel to make a lap if was 10 feet off the buoys the whole time on the course?

 

[Don Ferrette]

I think what you may not have accounted for is decceleration time thru the turns & the acceleration time back to 100 MPH.

We've only seen speed drops in & out of corners 5 to 6mph on radar, not 30.............. :blink:

 

[John Finch]

Hey John,

This is the math I was given to figure MPH for a 1 mi. 2-lap corase. 29.33 divide by time in sec. times 20 divide by 0.488 = MPH. I know it dosent sound right, but it really is. Doing this math a 2-lap time of 17.00 sec. = 70.708 MPH. I think what you may not have accounted for is decceleration time thru the turns & the acceleration time back to 100 MPH.

Don

Don,

The boat was had on radar at 98 coming out of the turns. We are talking between one and two seconds from buoy one to buoy 3. No time for decell and excelleration.

If the figures are set for a line drawn thru the buoys, it would be impossible to drive there. We are talking a full straightaway distance discrepancy in one lap!

 

[piper_chuck]

1246.6666 feet at 100mph over 8.5 seconds

ok, so the boat traveled 366 feet further than it had to. That's more than a full straightaway!

How about this. Can you figure out how far the boat would have to travel to make a lap if was 10 feet off the buoys the whole time on the course?

approx 942.7 feet. I wrote a spreadsheet to run any numbers you want. Give me a shout if you want to try more numbers. BTW, if my sheet is right, 8 second laps at 100 MPH would come from 81.7 foot radius turns.

 

[Scott Norris]

See, that's why Chuck works for Microsoft and I work on Cars.

 

[piper_chuck]

See, that's why Chuck works for Microsoft and I work on Cars.

Just cause I got lucky and won at Chesapeake, there's no reason to insult me! :angry: