Will Bogging Hurt MPH..?
What it says..?
If you launch at the track, then bog because the 2 step is too low, and then the car eventually picks up and goes can that hurt MPH..?
If you launch at the track, then bog because the 2 step is too low, and then the car eventually picks up and goes can that hurt MPH..?
What I'm saying is if you launch and the car goes immediately compared to you launching and the car bogging for a moment then picking back up and going would that affect overall MPH.
As lets say you just covered 10-20 feet of the track with the car bogging.
As lets say you just covered 10-20 feet of the track with the car bogging.
Yes it should. Your trap speed usually indicates what your car is capable of but think of if you rolled off the line, went a bit then took off you have less distance to accelerate so I would assume your speed would be lower in the end. This would be more suitable for the drag racing section though. Street tuned 2 step launch = garbage on a properly setup track your almost guaranteed to bog.
Well that's my logic too, but I have someone on another forum telling me that's not possible that bogging doesn't affect MPH.
My car was dyno tuned, and the 2 step was set at 8000.
I've also thought maybe tire pressure was a little too low, because I added 2 psi in each tire and went 0.2 quicker but the same MPH.
My car was dyno tuned, and the 2 step was set at 8000.
I've also thought maybe tire pressure was a little too low, because I added 2 psi in each tire and went 0.2 quicker but the same MPH.
Yes it will. If your motor is bogging, your car will not be as accelerating as quickly as it would be if it were not bogging. Let's take the first 20 feet as the distance and use the kinematics equation, Vf^2 = Vi^2 + 2ax, to show why:
where,
Vf = final velocity of car after a displacement of x feet
Vi = initial velocity(which is 0 since the car accelerates from a complete stop)
a = acceleration of car
x = displacement(20 feet).
Case 1(motor is bogging):
Vi = 0 ft/s
a = 20 ft/s^2
x = 20 ft
Vf^2 = 0^2 + 2*20(ft/s^2)*20(ft). Therefore Vf = 28.2 ft/s after 20 feet, which is 19.2 mph at 20 feet.
Case 2(motor is not bogging,thus quicker acceleration)
Vi = 0 ft/s
a = 30 ft/s^2
x = 20 ft
Vf^2 = 0^2 + 2*30(ft/s^2)*20(ft). Therefore Vf = 34.64 ft/s after 20 feet, which is 23.61 mph at 20 feet.
Assuming that the car is no longer bogging after travelling 20 feet, the car in case 1 and case 2 SHOULD be accelerating at the same rate for the rest of the drag race.
Using the same equation Vf^2 = Vi^2 + 2ax, we can calculate the final speed(trap speed).
Vi = 19.2 mph (28.2ft/s) in case 1. and 23.61 mph (34.64 ft/s) in case 2. Since it is a quarter mile the displacement is 1320 feet, but the car has already traveled 20 feet when it bogged. Thus the displacement , x = 1300 ft.
Case 1(car bogged at start)
Vf^2 = 28.2(ft/s)^2 + 2*20(ft/s^2)*1300ft, therefore, Vf = 256.2 ft/s or 174.68 mph.
or case 2
Vf^2 = 34.64(ft/s)^2 + 2*20(ft/s^2)*1300ft, therefore Vf = 262.7 ft/s or 179.11 mph.
These final speeds are not practical because the car will not accelerate that quickly or constantly for that matter. The point is I went into way too much detail explaining this and that it will make a difference in your final MPH.
where,
Vf = final velocity of car after a displacement of x feet
Vi = initial velocity(which is 0 since the car accelerates from a complete stop)
a = acceleration of car
x = displacement(20 feet).
Case 1(motor is bogging):
Vi = 0 ft/s
a = 20 ft/s^2
x = 20 ft
Vf^2 = 0^2 + 2*20(ft/s^2)*20(ft). Therefore Vf = 28.2 ft/s after 20 feet, which is 19.2 mph at 20 feet.
Case 2(motor is not bogging,thus quicker acceleration)
Vi = 0 ft/s
a = 30 ft/s^2
x = 20 ft
Vf^2 = 0^2 + 2*30(ft/s^2)*20(ft). Therefore Vf = 34.64 ft/s after 20 feet, which is 23.61 mph at 20 feet.
Assuming that the car is no longer bogging after travelling 20 feet, the car in case 1 and case 2 SHOULD be accelerating at the same rate for the rest of the drag race.
Using the same equation Vf^2 = Vi^2 + 2ax, we can calculate the final speed(trap speed).
Vi = 19.2 mph (28.2ft/s) in case 1. and 23.61 mph (34.64 ft/s) in case 2. Since it is a quarter mile the displacement is 1320 feet, but the car has already traveled 20 feet when it bogged. Thus the displacement , x = 1300 ft.
Case 1(car bogged at start)
Vf^2 = 28.2(ft/s)^2 + 2*20(ft/s^2)*1300ft, therefore, Vf = 256.2 ft/s or 174.68 mph.
or case 2
Vf^2 = 34.64(ft/s)^2 + 2*20(ft/s^2)*1300ft, therefore Vf = 262.7 ft/s or 179.11 mph.
These final speeds are not practical because the car will not accelerate that quickly or constantly for that matter. The point is I went into way too much detail explaining this and that it will make a difference in your final MPH.
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you really went all this way just to prove a point and in which you are still wrong.
kudos to you boeing for putting it into a mathematical equation.
like i said on the OTHER thread on the OTHER forum, you are confusing shortening the track minus 20 feet because you bogged failing to realize you are still moving if you bogged. your total mph is still going to be the same plus or minus a minimal amount of mph not a whole 8mph.
you bogged off the line causeing you to cut a bad 60ft correct? you also said it took like a second or so off your time right? so what is it? you bogged stop and got back in? or you bogged for a whole second and then got back in after 20ft? im tryn to figure this out for you because you still cut a 2.0 60ft dick.
kudos to you boeing for putting it into a mathematical equation.
like i said on the OTHER thread on the OTHER forum, you are confusing shortening the track minus 20 feet because you bogged failing to realize you are still moving if you bogged. your total mph is still going to be the same plus or minus a minimal amount of mph not a whole 8mph.
you bogged off the line causeing you to cut a bad 60ft correct? you also said it took like a second or so off your time right? so what is it? you bogged stop and got back in? or you bogged for a whole second and then got back in after 20ft? im tryn to figure this out for you because you still cut a 2.0 60ft dick.
I know the numbers are incorrect because i pulled an acceleration out of my a**, and those equations assume a CONSTANT acceleration which is also not going to happen. I highly doubt his civic will do 170+ mph lol. I did not shorten the track by twenty feet. I did assume that the car was only bogging for the first twenty feet, and not for the rest of the 1300 feet. 1300 + 20 = 1320 feet, which is a 1/4 mile. The point I was trying to make is that it WILL affect the final MPH, and I am glad that you agree with me on that. I was not trying to be accurate, just trying to show him why with numbers versus some complete BS that anyone can ramble on about. It may be a negligible difference, but it WILL make a difference in the final MPH.
I know the numbers are incorrect because i pulled an acceleration out of my a**, and those equations assume a CONSTANT acceleration which is also not going to happen. I highly doubt his civic will do 170+ mph lol. I did not shorten the track by twenty feet. I did assume that the car was only bogging for the first twenty feet, and not for the rest of the 1300 feet. 1300 + 20 = 1320 feet, which is a 1/4 mile. The point I was trying to make is that it WILL affect the final MPH, and I am glad that you agree with me on that. I was not trying to be accurate, just trying to show him why with numbers versus some complete BS that anyone can ramble on about. It may be a negligible difference, but it WILL make a difference in the final MPH.
he's trying to justify its poor performance by saying it bogged and stalled, did a backflip or some ****, thats why it ran 13s and only trapped 99mph every pass out. all i was pointing out to him was for 230whp in a 2300lb car he should be trapping alot more than 99mph no matter what. thats all.
Yes it will. If your motor is bogging, your car will not be as accelerating as quickly as it would be if it were not bogging. Let's take the first 20 feet as the distance and use the kinematics equation, Vf^2 = Vi^2 + 2ax, to show why:
where,
Vf = final velocity of car after a displacement of x feet
Vi = initial velocity(which is 0 since the car accelerates from a complete stop)
a = acceleration of car
x = displacement(20 feet).
Case 1(motor is bogging):
Vi = 0 ft/s
a = 20 ft/s^2
x = 20 ft
Vf^2 = 0^2 + 2*20(ft/s^2)*20(ft). Therefore Vf = 28.2 ft/s after 20 feet, which is 19.2 mph at 20 feet.
Case 2(motor is not bogging,thus quicker acceleration)
Vi = 0 ft/s
a = 30 ft/s^2
x = 20 ft
Vf^2 = 0^2 + 2*30(ft/s^2)*20(ft). Therefore Vf = 34.64 ft/s after 20 feet, which is 23.61 mph at 20 feet.
Assuming that the car is no longer bogging after travelling 20 feet, the car in case 1 and case 2 SHOULD be accelerating at the same rate for the rest of the drag race.
Using the same equation Vf^2 = Vi^2 + 2ax, we can calculate the final speed(trap speed).
Vi = 19.2 mph (28.2ft/s) in case 1. and 23.61 mph (34.64 ft/s) in case 2. Since it is a quarter mile the displacement is 1320 feet, but the car has already traveled 20 feet when it bogged. Thus the displacement , x = 1300 ft.
Case 1(car bogged at start)
Vf^2 = 28.2(ft/s)^2 + 2*20(ft/s^2)*1300ft, therefore, Vf = 256.2 ft/s or 174.68 mph.
or case 2
Vf^2 = 34.64(ft/s)^2 + 2*20(ft/s^2)*1300ft, therefore Vf = 262.7 ft/s or 179.11 mph.
These final speeds are not practical because the car will not accelerate that quickly or constantly for that matter. The point is I went into way too much detail explaining this and that it will make a difference in your final MPH.
where,
Vf = final velocity of car after a displacement of x feet
Vi = initial velocity(which is 0 since the car accelerates from a complete stop)
a = acceleration of car
x = displacement(20 feet).
Case 1(motor is bogging):
Vi = 0 ft/s
a = 20 ft/s^2
x = 20 ft
Vf^2 = 0^2 + 2*20(ft/s^2)*20(ft). Therefore Vf = 28.2 ft/s after 20 feet, which is 19.2 mph at 20 feet.
Case 2(motor is not bogging,thus quicker acceleration)
Vi = 0 ft/s
a = 30 ft/s^2
x = 20 ft
Vf^2 = 0^2 + 2*30(ft/s^2)*20(ft). Therefore Vf = 34.64 ft/s after 20 feet, which is 23.61 mph at 20 feet.
Assuming that the car is no longer bogging after travelling 20 feet, the car in case 1 and case 2 SHOULD be accelerating at the same rate for the rest of the drag race.
Using the same equation Vf^2 = Vi^2 + 2ax, we can calculate the final speed(trap speed).
Vi = 19.2 mph (28.2ft/s) in case 1. and 23.61 mph (34.64 ft/s) in case 2. Since it is a quarter mile the displacement is 1320 feet, but the car has already traveled 20 feet when it bogged. Thus the displacement , x = 1300 ft.
Case 1(car bogged at start)
Vf^2 = 28.2(ft/s)^2 + 2*20(ft/s^2)*1300ft, therefore, Vf = 256.2 ft/s or 174.68 mph.
or case 2
Vf^2 = 34.64(ft/s)^2 + 2*20(ft/s^2)*1300ft, therefore Vf = 262.7 ft/s or 179.11 mph.
These final speeds are not practical because the car will not accelerate that quickly or constantly for that matter. The point is I went into way too much detail explaining this and that it will make a difference in your final MPH.
correct.. with the right numbers where he pulled out of his ***..
man
thats all wel and good and thanks for that. what you missed in the other thread was the fact that his car make 230whp and bare gets into the 13s and traps 99 mph every time.
he's trying to justify its poor performance by saying it bogged and stalled, did a backflip or some ****, thats why it ran 13s and only trapped 99mph every pass out. all i was pointing out to him was for 230whp in a 2300lb car he should be trapping alot more than 99mph no matter what. thats all.
he's trying to justify its poor performance by saying it bogged and stalled, did a backflip or some ****, thats why it ran 13s and only trapped 99mph every pass out. all i was pointing out to him was for 230whp in a 2300lb car he should be trapping alot more than 99mph no matter what. thats all.
My car never once made 230whp on the dyno, so now you feel the need to lie since you were wrong. Lmao
Everytime..? I've been to the track twice with the car both hot summer days lmfao. And, I've made 8 passes and that's it, 3 of which were aborted to a missed gear.
Again I never made 230whp, and my car weighs more than 2300 pounds without me in it lmfao, and I'm 6 3 225 pounds.
You were wrong, talked all that ****, and now you are saying you were just trying to prove to me my car should go faster at it's current power level which I already know it should, but again try less than 230whp and about 2600 pounds with driver. It's funny to see you back pedaling though lol, so man up and admit you were wrong.
So, like you said in the other thread... How about you get on my level since I was right and you were wrong. Or how about who's foot is down who's throat now LMFAO
Last edited by B20vEf9; Dec 27, 2010 at 03:55 PM.
I know the numbers are incorrect because i pulled an acceleration out of my a**, and those equations assume a CONSTANT acceleration which is also not going to happen. I highly doubt his civic will do 170+ mph lol. I did not shorten the track by twenty feet. I did assume that the car was only bogging for the first twenty feet, and not for the rest of the 1300 feet. 1300 + 20 = 1320 feet, which is a 1/4 mile. The point I was trying to make is that it WILL affect the final MPH, and I am glad that you agree with me on that. I was not trying to be accurate, just trying to show him why with numbers versus some complete BS that anyone can ramble on about. It may be a negligible difference, but it WILL make a difference in the final MPH.
Wow so I was right the whole time and boeing just proved that, so how does it feel to be wrong like always..? Lmfao
My car never once made 230whp on the dyno, so now you feel the need to lie since you were wrong. Lmao
Everytime..? I've been to the track twice with the car both hot summer days lmfao. And, I've made 8 passes and that's it, 3 of which were aborted to a missed gear.
Again I never made 230whp, and my car weighs more than 2300 pounds without me in it lmfao, and I'm 6 3 225 pounds.
You were wrong, talked all that ****, and now you are saying you were just trying to prove to me my car should go faster at it's current power level which I already know it should, but again try less than 230whp and about 2600 pounds with driver. It's funny to see you back pedaling though lol, so man up and admit you were wrong.
So, like you said in the other thread... How about you get on my level since I was right and you were wrong. Or how about who's foot is down who's throat now LMFAO
My car never once made 230whp on the dyno, so now you feel the need to lie since you were wrong. Lmao
Everytime..? I've been to the track twice with the car both hot summer days lmfao. And, I've made 8 passes and that's it, 3 of which were aborted to a missed gear.
Again I never made 230whp, and my car weighs more than 2300 pounds without me in it lmfao, and I'm 6 3 225 pounds.
You were wrong, talked all that ****, and now you are saying you were just trying to prove to me my car should go faster at it's current power level which I already know it should, but again try less than 230whp and about 2600 pounds with driver. It's funny to see you back pedaling though lol, so man up and admit you were wrong.
So, like you said in the other thread... How about you get on my level since I was right and you were wrong. Or how about who's foot is down who's throat now LMFAO
im not back pedaling because what i said still stands true. what boeing did with the math was great and it proved exactly what i was telling you from jump.you just have to put 2&2 together to see it. everybody else does. a "bog" into a bad 60ft then continue the run accelerating and it will still have the same trap speed = weight/ horsepower. boeings equation gives a set consistant acceleration (but you shift gears and come off the throttle) which is not the case we have been debating dick. you are the one rewording it to make it seem as if you "won" this arguement. whatever floats your boat bitchboy. im not gonna waste anymore time beating a dead horse. fix the bogging issuse and you will see your ET drop but mph will be the same.
The car is now retuned on corn fuel.
Let me start off by saying IMW's dyno had a previous on going issue that was finally fixed which I wasn't aware of. Well with this new fix came a software update which now takes the dyno from reading like a dyno jet to reading like a mustang dyno(You will see what I mean) lol
The car ended up making 190whp on 93 octane(previously 216whp)
The car made 200whp on e85, so I guess with a solid 10hp gain you could say it would have made 226 if the same correction was applied as my first tune, but **** it I don't care what it makes on the dyno I care what it runs on the track.
Let me start off by saying IMW's dyno had a previous on going issue that was finally fixed which I wasn't aware of. Well with this new fix came a software update which now takes the dyno from reading like a dyno jet to reading like a mustang dyno(You will see what I mean) lol
The car ended up making 190whp on 93 octane(previously 216whp)
The car made 200whp on e85, so I guess with a solid 10hp gain you could say it would have made 226 if the same correction was applied as my first tune, but **** it I don't care what it makes on the dyno I care what it runs on the track.
97 civic dx coupe with full interior.
The car clearly has more in it as I pulled a car on my friends supercharged gsr hatch that trapped 103mph consistently and that was before my 3 inch exhaust and e85.
The car clearly has more in it as I pulled a car on my friends supercharged gsr hatch that trapped 103mph consistently and that was before my 3 inch exhaust and e85.
That hatch is probably 300 pounds lighter and I made 216whp on 93 at imw, and when I was retuned there on e85, dynapack sent a new firmware or software update which made the dyno read even lower. I made 200whp on e85 that day.
To answer your original question, NO.
Go to a track turn off your 2 step. Try launching at different rpms. I know this will be kind of hard, 4000, 5000, 6000, 7000, 8000. Your MPH on each of these runs, should be within 1mph if you are a consistent driver, don't miss a shift. The ET on the other had should be changing, depending on how much you bog and spin the tires.
Some of your numbers are just not lining up very well.
My Integra 2700lbs + me 200lbs at 170/130 dynojet ran 98-99mph. Now my Integra goes through the 3rd-4th shift at 105-106mph at the 190 power level. My typical shift point at the end of the 1/4 mile. I don’t have official track numbers yet.
I know dynopacks read high comparably speaking. If they did an update to the dynopack software and it now reads about 200whp that makes a bit more sense.
What rpm are you shifting at?
Is that shift point 200-300 rpm past you peak HP?
Does your torque curve have any excessive dips in power that your shift point could be landing in?
Your not running 17” at the track are you?
Go to a track turn off your 2 step. Try launching at different rpms. I know this will be kind of hard, 4000, 5000, 6000, 7000, 8000. Your MPH on each of these runs, should be within 1mph if you are a consistent driver, don't miss a shift. The ET on the other had should be changing, depending on how much you bog and spin the tires.
Some of your numbers are just not lining up very well.
My Integra 2700lbs + me 200lbs at 170/130 dynojet ran 98-99mph. Now my Integra goes through the 3rd-4th shift at 105-106mph at the 190 power level. My typical shift point at the end of the 1/4 mile. I don’t have official track numbers yet.
I know dynopacks read high comparably speaking. If they did an update to the dynopack software and it now reads about 200whp that makes a bit more sense.
What rpm are you shifting at?
Is that shift point 200-300 rpm past you peak HP?
Does your torque curve have any excessive dips in power that your shift point could be landing in?
Your not running 17” at the track are you?






