Math for fun and racing!
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From: VillePlatte, Louisiana
Don, I tried with a regular calculator, using the directions you gave in your example. I'm missing something because I cannot get the cube root of 6. I tried running my trucks numbers 13.95 sec, 103 mph, weight is 7000 lbs. I'm comming up with any where from 500 to 600 hp. I'm baffled 
Of course I have trouble getting my fuel mileage right. 
Regards,
Gene
Of course I have trouble getting my fuel mileage right. Regards,
Gene
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From: In the Shop
Ok, here it is:
Since you have the ET and the weight, we wont need a cube root but a cubed number. Its easier.
HP = ET/5.825
13.95/5.825 = 2.3948
cube it as follows: 2.3948 x 2.3948 x 2.3948 = 13.734
Now take the weight of 7000 and divide it by your cubed number: 7000/13.73 = 509.83 HP!
Now, if the truck dynos less than this we know the constant of 5.825 is too generous and certainly should be because the normally aspirated gas engine is way differnt than the TurboDiesel. Also, you may have 4 wheel drive and this can aid in traction VS a NA gasser with 2WD.
You could use my constant of 5.425 as a starting place.
13.95/5.425 = 17.00
7000/1700 = 411 HP.
Don~
Since you have the ET and the weight, we wont need a cube root but a cubed number. Its easier.
HP = ET/5.825
13.95/5.825 = 2.3948
cube it as follows: 2.3948 x 2.3948 x 2.3948 = 13.734
Now take the weight of 7000 and divide it by your cubed number: 7000/13.73 = 509.83 HP!
Now, if the truck dynos less than this we know the constant of 5.825 is too generous and certainly should be because the normally aspirated gas engine is way differnt than the TurboDiesel. Also, you may have 4 wheel drive and this can aid in traction VS a NA gasser with 2WD.
You could use my constant of 5.425 as a starting place.
13.95/5.425 = 17.00
7000/1700 = 411 HP.
Don~
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From: In the Shop
Gene,
Olee Poole had a full size B-1 and M4's. He dynoed on a DynoJet @ 522 HP. That is a set up pretty close to what you have.
I suspect the 509 we got with the formula using Patricks numbers are pretty close.
Turbo spool and lag plays a big part in ET for these engines and what they show on the DynoJet.
Don~
Olee Poole had a full size B-1 and M4's. He dynoed on a DynoJet @ 522 HP. That is a set up pretty close to what you have.
I suspect the 509 we got with the formula using Patricks numbers are pretty close.
Turbo spool and lag plays a big part in ET for these engines and what they show on the DynoJet.
Don~
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From: Cummins Technical Center, IN
Well, wouldn't you know it! I happen to have my trusty TI-89 right by my side! Want me to graph the Schied rail vs Bentz vs Keating?? HAHAHA.
Don, the tweaking of constants is the key, as you mentioned. For a 6-speed truck like mine, the constant would probably have to be closer 6 even.
Now, I've dynoed a 390 (dynojet, and very fishy), and a 350 (superflow, probably closer) in corrected numbers.
Given my weight of 6800#, let's take the 390hp figure. Assuming I don't miss a shift, Hale's constant puts me at a 15.10 ET. Not likely at all.
More realistic, imo, would be the 350hp number with a constant of 6 even. That puts me at 16.13ET-- much more realistic, imo.
This given that I had DD2s and an EZ.
If you have a hard time finding the 3rd root, you can also try raising the number to the "1/3rd" power. It's mathematically the same.
Also, if you are running a Microsoft Operating System, then you can just use the calculator that comes with it. Change the "view" to scientific, and you have all you need. To do a cube root, just check the "inv" box, then use the "X^3" key, key because you are doing the inverse of cubing-- you're taking the cube root.
Don, the tweaking of constants is the key, as you mentioned. For a 6-speed truck like mine, the constant would probably have to be closer 6 even.
Now, I've dynoed a 390 (dynojet, and very fishy), and a 350 (superflow, probably closer) in corrected numbers.
Given my weight of 6800#, let's take the 390hp figure. Assuming I don't miss a shift, Hale's constant puts me at a 15.10 ET. Not likely at all.
More realistic, imo, would be the 350hp number with a constant of 6 even. That puts me at 16.13ET-- much more realistic, imo.
This given that I had DD2s and an EZ.
If you have a hard time finding the 3rd root, you can also try raising the number to the "1/3rd" power. It's mathematically the same.
Also, if you are running a Microsoft Operating System, then you can just use the calculator that comes with it. Change the "view" to scientific, and you have all you need. To do a cube root, just check the "inv" box, then use the "X^3" key, key because you are doing the inverse of cubing-- you're taking the cube root.
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From: McKinney, TEXAS
Originally Posted by HOHN
Well, wouldn't you know it! I happen to have my trusty TI-89 right by my side!
I hate RPN (ala Hewlett Packard)
Back to your regularly scheduled thread,
~Rob
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From: In the Shop
Thats funny I have a TI-89 too.
Certainly some tweaks are needed and a stick truck will need major tweaks because the boost typically falls off between shifts. The average HP is much lower for the stick truck. Its the average that really matters anyway, not the peak, but we dont want to give away everything. LOL
Thanks for the tip on the MS calc. I never use that thing, but it seems the numbers are easier to get than using my million button method.
Now if we are all silent like Joe would like, we would not have learned this from the discussion. Joe (TS actually) hides his real numbers with inflated weights, 4WD traction and Nitrous that keeps the average HP higher by virtually eliminating turbo lag.
Lets look at Joes by the numbers:
10.97/5.825 = 1.8832
cubed 1.8832 = 6.6786
7000/6.6786 = 1048 HP!
Several things are easily tweaked to fix the disparity between a VP44 truck and a P pumped 12V monster.
Firstly we know there is very little turbo lag with a Nitrous shot feeding it. Secondly the reported weight could be inaccurate. I heard the scale numbers were closer to high 6700. Finally the 4WD has little tire slip.
Lets use a real world number like my constant of 5.425
10.97/5.425 = 2.0221
cubed = 8.2681
7000/8.2681 = 846 HP
Thats more like it. Or perhaps play with the weight some.
Lets try 6800. That gives 822HP. Or the reported 6700 gives 810. Thats getting more accurate even yet.
Or play with traction. Etc, etc.
Those are the peak numbers, but their average is high as well. Slow engine speeds yeild big HP with Nitrous VS just Diesel fuel and twin turbos.
Someone at the DHRA has been tweaking numbers to get to the drug assisted 6200 lb weigh VS the Diesel only 5500 pound weight restrictions for the Pro catagory next year. That same someone must have surely looked at their own numbers at the same time. Hint hint. Think California, dragsters and "I dont want to race it right now" LOL
Let use 6200 for Joes and see what happens if he does not change his HP.
We have the weight and the HP of say 850.
6200/850 = 1.9393 x the constant of 5.825 = 11.29
Well we know that constant is not good because it has run a 10.97.
Lets use mine to reconcile them backwards
1.9393 x 5.425 = 10.52 ET
Not bad and certainly competitive with the Garmon truck. That also jives with his current weight of less than 7000 lbs being very accurate.
BS is BS and numbers games are played by racers all the time. Math will always catch them. They cant hide their ET or MPH.
Use the formulas given above to reconcile their weight or HP from their MPH as well.
Notice they never dyno it in public? LOL
Don~
Certainly some tweaks are needed and a stick truck will need major tweaks because the boost typically falls off between shifts. The average HP is much lower for the stick truck. Its the average that really matters anyway, not the peak, but we dont want to give away everything. LOL
Thanks for the tip on the MS calc. I never use that thing, but it seems the numbers are easier to get than using my million button method.
Now if we are all silent like Joe would like, we would not have learned this from the discussion. Joe (TS actually) hides his real numbers with inflated weights, 4WD traction and Nitrous that keeps the average HP higher by virtually eliminating turbo lag.
Lets look at Joes by the numbers:
10.97/5.825 = 1.8832
cubed 1.8832 = 6.6786
7000/6.6786 = 1048 HP!
Several things are easily tweaked to fix the disparity between a VP44 truck and a P pumped 12V monster.
Firstly we know there is very little turbo lag with a Nitrous shot feeding it. Secondly the reported weight could be inaccurate. I heard the scale numbers were closer to high 6700. Finally the 4WD has little tire slip.
Lets use a real world number like my constant of 5.425
10.97/5.425 = 2.0221
cubed = 8.2681
7000/8.2681 = 846 HP
Thats more like it. Or perhaps play with the weight some.
Lets try 6800. That gives 822HP. Or the reported 6700 gives 810. Thats getting more accurate even yet.
Or play with traction. Etc, etc.
Those are the peak numbers, but their average is high as well. Slow engine speeds yeild big HP with Nitrous VS just Diesel fuel and twin turbos.
Someone at the DHRA has been tweaking numbers to get to the drug assisted 6200 lb weigh VS the Diesel only 5500 pound weight restrictions for the Pro catagory next year. That same someone must have surely looked at their own numbers at the same time. Hint hint. Think California, dragsters and "I dont want to race it right now" LOL
Let use 6200 for Joes and see what happens if he does not change his HP.
We have the weight and the HP of say 850.
6200/850 = 1.9393 x the constant of 5.825 = 11.29
Well we know that constant is not good because it has run a 10.97.
Lets use mine to reconcile them backwards
1.9393 x 5.425 = 10.52 ET
Not bad and certainly competitive with the Garmon truck. That also jives with his current weight of less than 7000 lbs being very accurate.
BS is BS and numbers games are played by racers all the time. Math will always catch them. They cant hide their ET or MPH.
Use the formulas given above to reconcile their weight or HP from their MPH as well.
Notice they never dyno it in public? LOL
Don~
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From: Jeffersonville, Ohio
Very informative discussion...this means that my 360HP, 8300lb truck will run through the 1/4 in 16.579s @ 82mph. Sounds about right to me...if I don't miss a shift and the DD doesn't slow me down too much!
Chris
Chris
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oh boy, time to have fun.
Here is the relationship between Horsepower and Torque, and a proof of where the constant 5252 came from.
Power, and Torque
W=work
P= power
Τ=torque
W=∫Τ dθ
dW= Τ dθ
P=dW/dt
P= Τ dθ/dt
ω=dθ/dt
P= Τ ω
One revolution is 2π radians
Thus
P= Τ ω
Becomes
P= Τ 2πr
HP=550 Ft*lbf/s
550 Ft*lbf/s)= Τ 2πr
60s=1min
(60s/min )(550 Ft*lbf/s)= Τ 2πr
Ft*lbf (550*60/2πr/ min) = Τ
Τ=HP(5252.11)/RPM
Here is the relationship between Horsepower and Torque, and a proof of where the constant 5252 came from.
Power, and Torque
W=work
P= power
Τ=torque
W=∫Τ dθ
dW= Τ dθ
P=dW/dt
P= Τ dθ/dt
ω=dθ/dt
P= Τ ω
One revolution is 2π radians
Thus
P= Τ ω
Becomes
P= Τ 2πr
HP=550 Ft*lbf/s
550 Ft*lbf/s)= Τ 2πr
60s=1min
(60s/min )(550 Ft*lbf/s)= Τ 2πr
Ft*lbf (550*60/2πr/ min) = Τ
Τ=HP(5252.11)/RPM
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From: In the Shop
Originally Posted by Diesel Freak
oh boy, time to have fun.
Here is the relationship between Horsepower and Torque, and a proof of where the constant 5252 came from.
Power, and Torque
W=work
P= power
Τ=torque
W=∫Τ dθ
dW= Τ dθ
P=dW/dt
P= Τ dθ/dt
ω=dθ/dt
P= Τ ω
One revolution is 2π radians
Thus
P= Τ ω
Becomes
P= Τ 2πr
HP=550 Ft*lbf/s
550 Ft*lbf/s)= Τ 2πr
60s=1min
(60s/min )(550 Ft*lbf/s)= Τ 2πr
Ft*lbf (550*60/2πr/ min) = Τ
Τ=HP(5252.11)/RPM
Here is the relationship between Horsepower and Torque, and a proof of where the constant 5252 came from.
Power, and Torque
W=work
P= power
Τ=torque
W=∫Τ dθ
dW= Τ dθ
P=dW/dt
P= Τ dθ/dt
ω=dθ/dt
P= Τ ω
One revolution is 2π radians
Thus
P= Τ ω
Becomes
P= Τ 2πr
HP=550 Ft*lbf/s
550 Ft*lbf/s)= Τ 2πr
60s=1min
(60s/min )(550 Ft*lbf/s)= Τ 2πr
Ft*lbf (550*60/2πr/ min) = Τ
Τ=HP(5252.11)/RPM
Don~
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Taking the guess work and “adjusted constants” out of the discussion
Let’s look at basic Physics.
Force = Mass x Acceleration
Work = Force x Distance
Power = Work / Time
Therefore,
Power = (Mass x Acceleration x Distance)/Time
Final Velocity = initial Velocity + Acceleration x Time
MPH=1.46 feet/second
Changing this to find average acceleration during a ¼ mile run, we have
(Trap speed (1.46))/ET = average Acceleration
Force exerted during the run F=ma
(Vehicle weight x Average Acceleration) 32 feet/second^2
Work (energy used during the run) = Fd = Force x 1320 feet
Average Power required to make the run = Work/time = Work/ET
1 Horsepower = 550 Foot pounds per second
Cramming all this together and neglecting air resistance
Average power required to make the ¼ mile run =
Weight of vehicle x 1.46 x (trap speed) x 1320 feet / (32 (ET)^2 x 550)
Finally,
HP= 0.1095(Weight x (Trap Speed))/ET^2
Let’s look at basic Physics.
Force = Mass x Acceleration
Work = Force x Distance
Power = Work / Time
Therefore,
Power = (Mass x Acceleration x Distance)/Time
Final Velocity = initial Velocity + Acceleration x Time
MPH=1.46 feet/second
Changing this to find average acceleration during a ¼ mile run, we have
(Trap speed (1.46))/ET = average Acceleration
Force exerted during the run F=ma
(Vehicle weight x Average Acceleration) 32 feet/second^2
Work (energy used during the run) = Fd = Force x 1320 feet
Average Power required to make the run = Work/time = Work/ET
1 Horsepower = 550 Foot pounds per second
Cramming all this together and neglecting air resistance
Average power required to make the ¼ mile run =
Weight of vehicle x 1.46 x (trap speed) x 1320 feet / (32 (ET)^2 x 550)
Finally,
HP= 0.1095(Weight x (Trap Speed))/ET^2
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From: In the Shop
Originally Posted by Diesel Freak
If anyone has the Aerodynamic drag coefficient of a dodge 2nd gen truck, then we can get down to the gnatts butt!!!
life is good
D~
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From: Greeley, Co.
Don,
I once heard someone say these diesel trucks weren't rocket science.
Your move to area 51 has taken you over the edge??? (I,m afraid)
Please keep us informed how things are there.
LaserBob
I once heard someone say these diesel trucks weren't rocket science.
Your move to area 51 has taken you over the edge??? (I,m afraid)
Please keep us informed how things are there.
LaserBob