oval speed calculation

[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?

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.

ok then.....how wide a turn would be needed to travel 1246 feet on the course for one lap if you shaved buoy two and 5?

 

[Don Ferrette]

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.

The turn radius for the IMPBA 1/3 mile oval is 35 feet, total distance 880 feet as listed on page K-12 of IMPBA rulebook. :blink:

 

[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?

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.

The turn radius for the IMPBA 1/3 mile oval is 35 feet, total distance 880 feet as listed on page K-12 of IMPBA rulebook. :blink:

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.

The turn radius for the IMPBA 1/3 mile oval is 35 feet, total distance 880 feet as listed on page K-12 of IMPBA rulebook. :blink:

yeah, but you can't drive thru the buoys. The center line of the boat has to be at least two feet to the outside of the centerline of the buoys.

 

[Scott Norris]

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:

You won Chesapeake cause you won!!! Break out the spread sheet Chuck. John wants to know how far he went at 100 MPH!!!

 

[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.

The turn radius for the IMPBA 1/3 mile oval is 35 feet, total distance 880 feet as listed on page K-12 of IMPBA rulebook. :blink:

Yup, I realize this. I set the spreadsheet up to allow me to input seconds per lap and calculate speed or input speed and calculate seconds per lap. I can then vary the turn radius for either and see what happens. Both variations come out the same way, if you're running 100 MPH and turning 8 second laps, your turn radius is 81.7 feet. It's possible I screwed up the math, but when I try 8.5 second laps at 35 foot radius, it comes out near 70 MPH.

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:

You won Chesapeake cause you won!!! Break out the spread sheet Chuck. John wants to know how far he went at 100 MPH!!!

It's telling me that an 8.5 second lap at 100 mph would be a turn radius of more than 93 feet!

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.

ok then.....how wide a turn would be needed to travel 1246 feet on the course for one lap if you shaved buoy two and 5?

Well, if you just did a big circle, I think the radius would be around 198 feet.

With all of us sitting here asking questions and posting replies, it seems this would be a good time to go to a chat, but didn't I read something about doing away with that option?

 

[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?

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.

The turn radius for the IMPBA 1/3 mile oval is 35 feet, total distance 880 feet as listed on page K-12 of IMPBA rulebook. :blink:

Yup, I realize this. I set the spreadsheet up to allow me to input seconds per lap and calculate speed or input speed and calculate seconds per lap. I can then vary the turn radius for either and see what happens. Both variations come out the same way, if you're running 100 MPH and turning 8 second laps, your turn radius is 81.7 feet. It's possible I screwed up the math, but when I try 8.5 second laps at 35 foot radius, it comes out near 70 MPH.

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:

You won Chesapeake cause you won!!! Break out the spread sheet Chuck. John wants to know how far he went at 100 MPH!!!

It's telling me that an 8.5 second lap at 100 mph would be a turn radius of more than 93 feet!

Thats only 10 feet off the front and rear straightawat buoy lines. Can you figure this............If I rubbed all the buoys, but did a football shape course and the appex in the middle of the straightaway was 30 feet off the buoy line. How far would the boat have traveled.

 

[piper_chuck]

It's telling me that an 8.5 second lap at 100 mph would be a turn radius of more than 93 feet!

Thats only 10 feet off the front and rear straightawat buoy lines.

I hope I don't find a silly little error in this spreadsheet. The turn radius is 35 feet, right? That means the number I quoted above is 58 feet off the bouy lines!

Can you figure this............If I rubbed all the buoys, but did a football shape course and the appex in the middle of the straightaway was 30 feet off the buoy line. How far would the boat have traveled.

I need to spend a couple minutes pondering this one. I know a way to approximate it, but I have to do a few calculations to see if I remember trig. If these paths get much more complicated I'm going to have to dig out some old math books.

 

[John Finch]

It's telling me that an 8.5 second lap at 100 mph would be a turn radius of more than 93 feet!

Thats only 10 feet off the front and rear straightawat buoy lines.

I hope I don't find a silly little error in this spreadsheet. The turn radius is 35 feet, right? That means the number I quoted above is 58 feet off the bouy lines!

Can you figure this............If I rubbed all the buoys, but did a football shape course and the appex in the middle of the straightaway was 30 feet off the buoy line. How far would the boat have traveled.

I need to spend a couple minutes pondering this one. I know a way to approximate it, but I have to do a few calculations to see if I remember trig. If these paths get much more complicated I'm going to have to dig out some old math books.

If you can answer this one...... You are the man. Things are starting to make sense. I have driven far off the straightaway line and still hugged the buoys, making it appear that it was a perfect run when in fact an an egg shaped course is what had been run. I just didn't think it was enough to make a 30 mph difference from calculated speed to actual speed.

 

[John Knight]

John,

There is another point to this that should not be overlooked. Even though you were hitting about 100 mph going into the turn the shame of it all is that the AVERAGE speed was 70. That means that in order to drop the average down to 70, what does the bottom number have to be to offset the 100? 45 to 50? Just does not seem to be right. We all know that the boat is not dropping 50% of it's speed in the turn especially when there is a radar reading of only a 5 mph drop coming out of the turn. Funny math for sure!! Hard to explain to a non-racer and to get them to believe.

Lesson for me is that if I ever do any time trials, I shall stick to SAW's and not get screwed by the math. Straight line from points A to B is hard to dispute.

John

 

[piper_chuck]

Can you figure this............If I rubbed all the buoys, but did a football shape course and the appex in the middle of the straightaway was 30 feet off the buoy line. How far would the boat have traveled.

I need to spend a couple minutes pondering this one. I know a way to approximate it, but I have to do a few calculations to see if I remember trig. If these paths get much more complicated I'm going to have to dig out some old math books.

John, I'm approximating here, but I don't see that adding more than something like 15 to 20 feet.

 

[DON MAHER]

OK, See if I get this right. A 2-lap corase is 1760 ft. on top of the bouys so for every ft. off the bouys you travel 12ft. per lap farther.Thats 24 ft. per 2-laps. OK at 100 mph you travel 146.6 ft. per sec. so lets say you drove 1ft. off the bouys, you traveled 1782ft. Now 1782ft. divided by 146.6ft. per sec.= 12.15 sec. or just over 6 sec. per lap. Now can you see how 4 or 5 more sec. will drop the speed?

Don

 

[John Finch]

If being just 10 feet off the buoys makes 30 mph difference.......Transfere that to heat rcaing. The possibilitys are limitless. Thanks for doing all the calculations. It's time to sit down with Mary and watch Vegas.

OK, See if I get this right. A 2-lap corase is 1760 ft. on top of the bouys so for every ft. off the bouys you travel 12ft. per lap farther.Thats 24 ft. per 2-laps. OK at 100 mph you travel 146.6 ft. per sec. so lets say you drove 1ft. off the bouys, you traveled 1782ft. Now 1782ft. divided by 146.6ft. per sec.= 12.15 sec. or just over 6 sec. per lap. Now can you see how 4 or 5 more sec. will drop the speed?

Don

That's some awesome numbers. Another way to look at it.

 

[piper_chuck]

If being just 10 feet off the buoys makes 30 mph difference.......Transfere that to heat rcaing. The possibilitys are limitless. Thanks for doing all the calculations. It's time to sit down with Mary and watch Vegas.

At 100 MPH, being 10 feet off the bouys adds .4 seconds to each lap. At a more leisurely 70 mph, it would add .6 seconds. One more way to look at it, if a boat that's 4 MPH slower (66) hugs the bouys, it's going to smoke a 70 MPH boat that's clearing the bouys by 10 feet. Bottom line, hug the bouys.

 

[Kez]

This is how I look at it:

Average speed = total distance travelled / time

Total distance travelled in one lap is 880ft = 0.166 mile

Time the boat took to do one lap = 8.5 sec = 0.00236 hour

Average speed = 0.166 mile / 0.00236 hour = 70.33 mph

My 2 cents..

Kez

 

[DON MAHER]

John,

Any way you look at your record is awsome!!!!!!

Don

 

[Don Ferrette]

This is how I look at it:

Average speed = total distance travelled / time

Total distance travelled in one lap is 880ft = 0.166 mile

Time the boat took to do one lap = 8.5 sec = 0.00236 hour

Average speed = 0.166 mile / 0.00236 hour = 70.33 mph

My 2 cents..

Kez

Kez you're right, the math adds up on that there is no doubt. But I think the point is it makes things so deceiving & that's why people see the avg. mph & think they can do it with a heat race boat. :lol:

 

[Marty Davis]

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.

John:

Here is something that should help you a lot with this question. I had the same question and it was answered by this little spreadsheet. There is a place to click on the excel spreadsheet on this past tech note that will download this little application. Looking at your time/speed, it is REALLY outdated

http://rcboat.com/lap.htm

Marty Davis

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.

 

[Wesleys Custom Graphics]

Hay, I hope nobodys got copyright on this. This is goog stuff! We've ran sprint cars for the last 10 years and I had these calculations for most of the tracks we ran. It's nice letting these brain children do the math for you. THANKS!

ps lane 1 is not always the fast way around the track. I know some of you brain boys will dispute this but that's OK........."Im OK , your OK" or something like that

 

[Marty Davis]

Hay, I hope nobodys got copyright on this. This is goog stuff! We've ran sprint cars for the last 10 years and I had these calculations for most of the tracks we ran. It's nice letting these brain children do the math for you. THANKS!

ps lane 1 is not always the fast way around the track. I know some of you brain boys will dispute this but that's OK........."Im OK , your OK" or something like that

No copyright, just helpful stuff.....

Marty Davis

 

[John Finch]

Interesting chart. I like the way it fills out the blanks for you. Question........Is this chart only good if you figure a straight line thru the traps buoy 6 to 1? I have found that the oval course is fastest when driven with neutral rudder on the boat, which equates to an egg shaped course. One of the secrets to a faster lap time is not using the rudder down the straights but rather letting the boat run it's natural line.