It has to be the 1/4 mile.At most tracks is hard to drive around the water box not enough room.If you cant keep all the tires out of the water your better to drive thru it then when your out of the box do a burn out to dry the tires.I agree stage shallow & leave on last amber.The guy in the neon is probabley waiting for green..Your reaction time wont change your ET because the timer doesnt start tell you leave the beam.But it has a lot to do with who wins the race...Good luck & whip that neon...Lew

 

Wow didn't know that the transmissions and transfer cases could hold together that well. Good to know that as long as the wheels stay straight I will be ok.

 

Transfer case... yes.


Stock transmission.... not so much.

Remember, the torque converter is the weakest link in the entire thing, regardless of 2wd or 4wd (although 4wd boosted launches are a bit harder on it). Repeated runs at the drag strip took a toll on my stock transmission. I've upgraded now, and my current trans is pretty much bullet proof. Drag racing in any form is really hard on the stock trans, so be careful.

 

Yeah my transfer case, transmission and torque converter were changed and upgraded less than 6 months ago so I should be good to go. The way I look at it all forms of racing take a toll on parts it is just part of racing. I won't be racing this truck too much I am about to be working on a formula SAE car to take care of my racing woes.

 

I beat the crap out of my truck after I got my first DP Tunes. I made 11 runs down the 1/8 mile strip here in both 2wd and 4wd. My 4wd runs were the best. It can certainly take it just like Curtis said.

I say bring that Neon on. I got something for it. I smoked many a ricer that night. Now I am faster.

 

Here's an example from my computer model for analyzing drag race performance. The following graph is for a tuned 7.3L SRW F350 weighing 7,200 lb, with a 3.73 diff, a 4R100 tranny, and 31.6" diameter 265/75R16 tires. The truck's RWHP vs RPM curve has a maximum value of 340 RWHP at 2,850 RPM, and the tranny shift points were chosen to be at the "optimum MPH speeds" so that the maximum possible RWHP gets applied to the track at each instant during the run. The track temperature is 70 F and the barometric pressure is the standard sea level value of 14.7 psi. Unlike online drag race calculators my model is exact and it includes the aerodynamic drag and rolling resistance characteristics for the specific truck.

There's a lot of interesting Physics involved in achieving the ETs and MPH trap speeds at the commonly used reference distances of 60, 330, 660, 1000, 1320 feet, and my computer model gives results for all of these distances. Since "reaction time" is dependent on the driver's skill I don't include that in my model. My results just calculate the ET's from the starting line to the finish lines for the various reference distances. Also my MPH trap speeds are the "instantaneous" speeds at which the truck hits the various finish lines. This means they're what you'd get using a "radar gun" speed measurement, whereas some tracks employ several closely spaced timing measurements and that gives an "average" trap speed.

In the graph the "green" RWHP Total curve gives total RWHP that's applied to the track. The "blue" RWHP Accel curve gives the RWHP that's available to accelerate the truck after subtracting the RWHP that's required for overcoming aerodynamic drag and rolling resistance from the total RWHP that's applied to the track. At higher MPH speeds you can see the dramatic effects of aerodynamic drag and rolling resistance by looking at the difference between the "green" RWHP Total curve and the "blue" RWHP Accel curve. At 70 MPH it only requires 65 RWHP to overcome aerodynamic drag and rolling resistance but at 90 MPH as you cross the 1/4 mile distance it requires 160 RWHP to overcome aerodynamic drag and rolling resistance.

The "pink" RPM curve is read on the left hand scale by multiplying by it 10. The "gold" acceleration curve is read on the right hand scale by dividing it by 1,000 to get the "g" number of the acceleration. You can read the trap speeds from the "red" curve using the right hand side distance scale and the lower MPH scale.

As you can see with everything assumed to be optimum during the run this model predicts a 1/4 mile ET of 14.271 sec and a 90.56 MPH trap speed for a 340 maximum RWHP. Do you know your RWHP? If you post all your split times I can estimate your RWHP. It's looking like you'll probably need more than a 340 maximum RWHP to beat the neon!

Fold the mirrors in to reduce aerodynamic drag. Remove the tail gate to save weight and possibly reduce aerodynamic drag. Running with less than a half tank of fuel saves 100 lb in weight. I didn't see any advice given regarding tire pressure? Higher pressure reduces rolling resistance but getting the best grip requires lower pressure so there's probably a compromise toward the lower end?

 

Ernest, is your model designed to be optimal, or to mimick real life? If you are trying to model a real life drag race with a Superduty, I think the calculations need some slight adjustments. Your model is too "generous" on the ET's and MPH for the given amount of HP. A PSD with those tire and weight estimates will need more than 400 rwhp to turn low 14's at over 90 mph. PSD's with 350 rwhp are pulling 15.0 ET's in the best conditions.

Off the top of my head, the graph is not correct for your shift patterns. I know you said you chose these shift points as "optimal", but it doesn't mimick an actual PSD shift pattern, as the drop in RPM's is too drastic between shifts, especially the 1-2 shift.

Again, I'm not sure if this is just an "ideal" model or if you are trying to mimick a real life drag race. If it's just an ideal model, then that's fine, ignore what I said.

You bring up a good point. Tire pressure is tricky, even for 4wd trucks. Those with higher HP trucks tend to start dropping tire pressure, even when launching in 4wd. But it's all an experimentation to try and find the sweet spot, and this all depends on tire size and power you are putting to the ground. Basically when you race, start dropping tire pressure 5 psi at a time until your times start to get worse, then bump it back up to get your best times. Then there's finding that perfect balance of pressure between the front and rear tires too. All of this takes time, and you'll need a portable air compressor handy. With a truck running the basic mods (chip/intake/exhaust), changing tire pressure will have a smaller net gain in times, and dropping too much will make the truck slower.

 

I'm sure this will be more enjoyable than talking about the infamous 6637 air filter! My model is designed to not only "mimick real life" but to actually duplicate real life with 100% accuracy right down to the nearest millisecond! My model includes every possible parameter that effects the end result including for example the specific aerodynamic drag area for each truck and how altitude, barometric pressure, and ambient temperature effect aerodynamic drag. Of course as with any computer model the trick is to get an accurate data set for the specific truck being modeled. How about I build a model for your truck that's so exacting you'll be able to look at a weather report and predict your ET before going to the track?

I used a SRW F350 as an example because I've got measured data for its aerodynamic drag and rolling resistance characteristics but these can be easily determined for any truck by making MPH versus TIME measurements while coasting in neutral from various initial MPH speeds. Your 4WD truck is a little higher than a SRW F350 so this increases its aerodynamic drag area and that increases the 1/4 mile ET of 14.271 sec predicted for a SRW F350.

My model is like none other because I calculate and plot on a MPH basis and I use the increase in the truck's kinetic ENERGY that results from applying a given RWHP to the track for a given TIME and not a FORCE based acceleration approach. By taking the truck's gearing and tire diameter into account my approach allows the truck's RWHP versus RPM curve to be directly related to the RWHP that's applied to the track at each specific MPH speed. My approach allows the race to be conveniently divided into three approximately equal MPH segments, 0-30 MPH launch, 30-60 MPH mid-range, and 60-90 MPH top-end where completely different "Physics" are at work for determining the ETs in each segment.

Here's an example of how I came up with the shift patterns but it's for a lower RWHP truck than the one in the drag race graph. You first need a RWHP versus RPM curve like the one shown below.



Then you convert the above RWHP versus RPM curve into a series of RWHP versus MPH curves for each gear. In the example below I assumed that the TCC was locked all the way down to 0 MPH which of course isn't real life but otherwise I need a model of TC slippage and I'm working on one of those.

For the gearing shown below I think you'd get the lowest ET by locking the TCC at 20 MPH and keeping it locked solid for the rest of the race! You'd need to get Jody to give you a special tune to do that and to make the tranny shift at exactly the MPH shift points indicated. Of course these "optimum" MPH shift points depend on the exact shape of the RWHP versus RPM curve, the gearing, and the tire diameter.

The laws of "Physics" say that you'll get the minimum possible ET by applying the maximum possible RWHP to the track at each and every MPH speed as you race toward the finish line and that's what an "optimum" shifting program needs to do. I'd have to model whether or not you'd get a lower ET by trying to fill in the gaps between gears with an unlocked TC and eating that extra drive line loss or by keeping the TCC locked up throughout the race and eating the drop in RWHP between gears?



My example graph gives a 60 ft ET that's more appropriate for a 4WD truck than for a 2WD truck. I call the "gold" curve the "g" number of the acceleration because that's what most people relate to the easiest. The "g" number is also numerically equal to u={tire FORCE}/{truck WEIGHT} where u is the coefficient of friction which determines how much RWHP can "stick" to the track during launch. By getting some 60 ft ETs for your truck using different tire pressures I can model how the coefficient of friction for your tires varies with inflation pressure!

I could also estimate the benefit of getting a special set of drag race tires with a softer compound so that the tread would literally "stick" to the track and then the friction force is dependent on the "shear strength" of the rubber! At 20 MPH in the example graph the instantaneous g=0.802=u and this is a coefficient of friction that's achievable with "street" tires which is why I think locking the TCC at 20 MPH will let you apply your full RWHP to the track at that speed.

Here's another example of the neat things you can do with a computer model. The chart below gives the ETs and MPH trap speeds for 4 cases. Case #1 is an "idealized" result in which the maximum 340 RWHP is applied to the track continuously throughout the entire run and there's no aerodynamic drag or rolling resistance, no tranny shifting to worry about, and the tires have a perfect grip and don't slip. This is kind of like dropping a weight in a vacuum where RWHP takes the place of gravity! Case #2 includes tranny shifting and tire slippage but not aerodynamic drag and rolling resistance. Case #3 is just like the "idealized" result but it does include aerodynamic drag and rolling resistance. Case #4 includes everything and it's the one in the example graph.

As you can see the tire slippage is what makes the value of Avg "g" @ 20 MPH go from about 5 g without slippage to about 1.3 g with slippage. If you look at 1000 ft and longer the ETs get progressively longer as you go from case #1 to case #4, and this is because the aerodynamic drag and rolling resistance limits the ETs and MPH trap speeds at higher MPH speeds.

At 660 ft and shorter the aerodynamic drag and rolling resistance is less important than tire slippage as is indicated by the lower ET for case #3 versus for case #2! On the right-hand-side of the chart I compare the differences between case #4 and cases #1, #2, and #3, to illustrate the relative effects of shifting, tire slippage, and aerodynamic drag and rolling resistance.

If you look at #4 vs #3 you see that even my assumed best case "real world" shifting strategy increases the ET by 0.354 sec and I'll bet that most drivers actual shifting strategies increase their ET at least an additional 0.35 sec on top of that for a net increase in ET of at least 0.7 sec!

Keep in mind that NASCAR and F-1 teams pay engineers like myself 6 figure salaries to analyze stuff like this but I do it for free! However if the stock market doesn't eventually recover I might have to start charging a slight fee as well just to cover the wear and tear on my computer!

 

Ok, I see where you are coming from now. And I think I see why you calculated a 340 rwhp SRW F-350 would run a 14.2 ET. There are some things missing from the model, and even you pointed it out:
I can see why it's hard to come up with an accurate way to calculate ET's and trap speeds.

FWIW, when drag racing, the RPM's on a PSD will stay above 2500 after you leave the starting lights, and pretty much all the way down the track for every gear, including the T/C lockup. This includes a stock 4r100 as well as a modified 4r100.

 

Another thing my model can be used for is analyzing post race results. For example if you use AE to data log MPH and RPM, for each run and give me that along with all the ETs and trap speeds for each segment I can calculate the exact RWHP you applied to the track during each segment. If you also log BP and MAT I can get a fairly accurate estimate your FWHP during each segment. Of course I'd also need track temp, BARO pressure, and any wind adjustments.

 

I would like to know your opinion on barometer inchs & drag racing...I know about wind & track temp..But we were at the track about a mounth ago on a cool night & everone was running a few 10ths slower than usaual..When i would bring the truck in after a run the condesation was runing of the roof & down the back window after a run..Also before we ran there was some pro stock cars testing before they opened the track to us...They use a different compond tire..Just trying to learn some new info..Lew

 

Here's the "Physics" which explain your above observations. Having a higher barometer reading and or a cooler ambient air temperature increases the density of the air. While denser air going into your engine has the potential to produce more FWHP driving through a mass of denser air uses up this extra FWHP to overcome the increased aerodynamic drag that denser air causes and it usually requires some additional FWHP to overcome this increased aerodynamic drag above and beyond any extra FWHP your engine might generate due to the denser air!

The RWHPd that a truck has to generate to overcome aerodynamic drag is given by...

RWHPd={(AD)(Ad)(MPH^3)}/{(11,217.7)} HP

...the AD=Air Density lb/ft^3 is given by AD={(2.70325)(AAP)}/{(AAT+459.67)} lb/ft^3 where AAP=Atmospheric Air Pressure psi depends on the barometer reading and AAT=Atmospheric Air Temperature F. The Ad=Aerodynamic Drag Area ft^2 is given by Ad={(Cd)(Ap)} ft^2 where Ap=Projected Frontal Area ft^2 of the truck which is about its height x width and Cd=Drag Coefficient which is a dimensionless ratio that takes into account the truck's body shape.

According to my estimates the Ap=Projected Frontal Area ft^2 of an F350 is Ap=37.8 ft^2 and the F350 body shape has a Cd=Drag Coefficient of Cd=0.75, and this gives an Ad={(Cd)(Ap)}={(0.75)(37.8)}=28.35 ft^2.

Lets assume a "cool night" is an AAT=40 F and that you're at sea level so AAP=14.7 psi. This gives an AD={(2.70325)(AAP)}/{(AAT+459.67)}={(2.70325)(14.7)}/{(40+459.67)}=0.0795 lb/ft^3 so that as you're nearing the end of the 1/4 mile run at about 80 MPH the RWHPd={(AD)(Ad)(MPH^3)}/{(11,217.7)}={(0.0795)(28.35)(80^3)}/{(11,217.7)}=102.9 HP.

On a warm summer day at AAT=90 F the AD={(2.70325)(AAP)}/{(AAT+459.67)}={(2.70325)(14.7)}/{(90+459.67)}=0.0723 lb/ft^3 and at 80 MPH the RWHPd={(AD)(Ad)(MPH^3)}/{(11,217.7)}={(0.0723)(28.35)(80^3)}/{(11,217.7)}=93.5 HP.

So at 80 MPH it takes about 9.4 more RWHP to overcome aerodynamic drag at 40 F versus 90 F. The amount of extra FWHP you get at lower AAT depends on your tuning and on your IAD=Inlet Air Density lb/ft^3 which is the air density going into the air filter inlet . If you use an "open element" air filter your IAD depends on your engine compartment temperature and not directly on the AAT so you might not get much extra FWHP at lower AAT.

The graph below gives an "overview picture" of the RWHP that's required to overcome both aerodynamic drag plus the rolling resistance for a number of different trucks. You can see how the required RWHP for the F350 varies as a function of AAT and AAP. At higher MPH speeds the effect of higher aerodynamic drag at lower AAT or higher AAP will dominate any increases in FWHP at lower AAT or higher AAP because for a given AD the RWHP that's required to overcome aerodynamic drag increases as MPH^3!

High humidity reduces AD but only very slightly compared to dry air and the effect is so small and so complicated that I don't include it in my model. However a "dripping wet truck" will have a higher Cd than for a dry slick truck and this effect might cause an even larger increase in aerodynamic drag than a lower AAT does! The AAT or rather the actual track temperature effects tire grip in a complicated way in which too high or too low a tire temperature reduces tire grip.

 

Thanks Ernest !! Thats what i was looking for ..Lew

 

You asked about " barometer inches" and I forgot to give you the conversions to psi which you need to do in order to use my equation for AD which has AAP in psi.

For a track at sea level (0 ft altitude) on a "standard day" the exact barometer reading will be 29.921" Hg and that corresponds to an AAP=14.696 psi atmospheric pressure. This means that 1 psi=2.036" Hg and that 1" Hg=0.491 psi. So you divide the track's barometer reading by 2.036 to get the AAP in psi to use in my equation for AD to get the RWHP required for aerodynamic drag. So AAP={track barometer reading}/{2.036} psi.

Say you now go to a track at an 1,800 ft altitude and it's also a "standard day" there. The actual barometer reading at this track will be 28.12" Hg but if you call the local airport weather service they'll report the barometric pressure as 29.92" Hg because they've correct the track's actual 28.12" Hg barometric pressure on a "standard day" to what it would be if the track were at a 0 ft altitude on a "standard day"!

The airport does this so that a pilot like me can adjust his altimeter setting to a value of 29.92" Hg so that when his altimeter which is just a sensitive barometer that's calibrated in altitude reads the actual barometric pressure of 28.12" Hg it will calculate the correct altitude of 1,800 ft from the equation altitude ft={1,000}{(altimeter setting)-(barometer reading)} so that the altimeter reads an altitude ft={1,000}{(29.92)-(28.12)}={1,000}{1.80}=1,800 ft.

According to this above "altimeter formula" for each 1,000 ft increase in altitude above sea level (0 ft altitude) there's a 1" Hg decease in the reading of an actual barometer but this is only an approximation that's based on a "standard model" for how pressure varies with altitude.

Sometimes atmospheric conditions don't fit this "standard model" and this approximation has caused pilots to fly into a hill during their descent for a landing! Of course some "novice" pilots forget to set their altimeter or momentarily forget that their altimeter reads the number of ft above sea level and not the number of ft above ground level!

Say you're at this same track at an 1,800 ft altitude on a "non standard day" where a high pressure front is passing through and you call the local airport weather service and they report the barometric pressure as 30.13" Hg. To account for your known 1,800 ft altitude you need to subtract 1.80" Hg to get 30.13-1.80=28.33" Hg and an actual barometer at the track will read 28.33" Hg on this "non standard day" and you then get the track's AAP from AAP={track barometer reading}/{2.036}={28.33}/{2.036}=13.915 psi and then use this AAP in the equation for AD.

The equation to calculate the exact AAP for a "standard day" at different altitudes above sea level is complicated but you can use the graph below to read the "standard day" AAP as a function of altitude. Since high or low pressure fronts can easily change the actual the barometric pressure at any given location by plus or minus 1" Hg =1,000 ft you can also use this graph to estimate this effect.



In the graph below you can see how at a given altitude the "standard day" AAP and the AAT combine to determine the "standard day" AD at that altitude and how this AD varies as a function of AAT and altitude. This is the AD that's used in my equation for aerodynamic drag. You might wonder why the "standard day" AAP at the track doesn't depend on AAT and I plan on doing a post on air flow and air pressure where I'll try to explain that.