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Math majors, trig question!
#1633247
06/13/14 11:25 PM
06/13/14 11:25 PM
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gregsdart
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If you know the height and position of the CG, it should be possible to calculate just how much force it will take to overcome front end weight without the aid of springs? By that I mean a solid front could be lifted by XX percentage of the actual weight on the front, based on the math. The higher the cg, the less it will take; anybody know the formula?
8..603 156 mph best, 2905 lbs 549, indy 572-13, alky
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Re: Math majors, trig question!
[Re: jcc]
#1633249
06/14/14 01:26 AM
06/14/14 01:26 AM
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gregsdart
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There is no time component for the answer to the question, just a percentage value. Adding in the springs effect can be done later, if desired. For sure everything is changing
once the car moves, but it helps to have an idea of where things are at at the start.
Lets say the Cg was high enough to effect the needed force to lift the front by 6 percent, or some other number. A good chassis designer would be able to design the car around the intended ET, and launch torque for best results, rather than ending up with too violent of a car like I did due to setting my motor back too far or having the Cg too high, or both.
8..603 156 mph best, 2905 lbs 549, indy 572-13, alky
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Re: Math majors, trig question!
[Re: gregsdart]
#1633250
06/14/14 11:00 PM
06/14/14 11:00 PM
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No math major or trig needed here. What you're trying to do is rotate the car about the rear axle. Just set the weight times the horizontal distance from the CG to the rear axle equal to the unknown force times the vertical height of the CG MINUS the height of the axle, and solve for the unknown. weight x length 1 = Force x (CG Height - axle height) Force = (weight x L1) / L2 where L1 is the distance from CG to axle L2 is the height of CG minus axle height Now us good ol' F=MA to determine what acceleration is needed to generate that force. 
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Re: Math majors, trig question!
[Re: @#$%&*!]
#1633251
06/15/14 12:43 AM
06/15/14 12:43 AM
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gregsdart
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As the angle of the tire contact patch to the CG increases, the force needed to lift the car gets less and less due to leverage. But the change is not linear, or the same per degree of height change in the CG.
It will be much less change for the first degree versus say from 30 degrees to 31 degrees. That is where the trig comes in.
Last edited by gregsdart; 06/15/14 12:49 AM.
8..603 156 mph best, 2905 lbs 549, indy 572-13, alky
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Re: Math majors, trig question!
[Re: gregsdart]
#1633252
06/15/14 06:15 AM
06/15/14 06:15 AM
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Joined: Oct 2012
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Bridgewater, NJ
HT413
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You guys don't have it quite right. If you're doing the lifting at launch, remember first off all that your engine torque is multiplied by the rear ratio, then by the 1st gear ratio and also by the torque converter multiplication. Don't forget the torque is technically reduced by the larger size tire too (which is in part why using too tall a tire hurts your et's.)
Here's the deal:
Engine torque needed to lift the front end = (front end weight)(wheelbase)(tire radius) / (vertical height of cg above floor)(rear ratio)(1st gear ratio)(torque converter multiplication)
Note that all dimensions are in FEET, even tire diameter. Also, this doesn't count drivetrain losses.
Example:
Front weight is 1500lb
Wheelbase = 116" = 9.67' (69 roadrunner)
30" tire = 15" radius = 1.33' radius
2.45:1 1st gear (727 standard gearset)
4.56's
2:1 torque converter multiplication (some guys say as much as 2.5)
18" center of gravity above the ground (this is a guess) = 1.5'
Engine torque = (1500lb)(9.67')(1.33') / (1.5')(2.45)(4.56)(2)
Engine torque "at the rear wheels" = 563 ft-lbs
Add in your estimated drivetrain loss and this is the torque your motor needs to make to lift tires off the ground aside from chassis tuning. Remember, this is a static analysis, so chassis tuning can reduce this significantly as it helps to throw the weight onto the back tires essentially reducing the front end weight in the equation above. Also, this assumes perfect traction.
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Re: Math majors, trig question!
[Re: HT413]
#1633253
06/15/14 09:39 AM
06/15/14 09:39 AM
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gregsdart
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Quote:
You guys don't have it quite right. .
I am familiar with the rest of the math. All I want is the value that comes from the raised Cg. Because it is variable for each added degree of height, it can't be figured with simple math.
It is the same equation to the speed differential of a U joint. As the angle gets higher, the speed of the driven side of the joint speeds up and slows down in relation to the drive side. That is why simple U joints are always at 90 degrees to each other. The drive shaft itself speeds up and slows down at high angles, but the joints being 90 degrees apart cancel each other, so the output joint ends up as a perfect steady speed match to the input end.
8..603 156 mph best, 2905 lbs 549, indy 572-13, alky
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Re: Math majors, trig question!
[Re: gregsdart]
#1633254
06/15/14 10:18 AM
06/15/14 10:18 AM
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gregsdart
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I found a formula that measures the value using simple math combined with a mechanical drawing. This formula is used to figure lateral loading and unloading of the outside tires versus the inside tires in a turn, to figure how much total loss of traction happens due to uneven loading.
With the ground being perfectly horizontal, the higher you raise the Cg the less input torque required to change the percent of load change.
If you measure at several points, and divide the length of the dashed line by the length of the horizontal line (the radious) we get the answer!
it shows that the first couple of degrees the percent is very close, at 45 degrees it is at the biggest gain per degree, and continuing towards 90 (straight up) it decreases at the same rate it gained by degree.
Having trouble finding the scan . It is on page four of
CHASSIS ENGINEERING by Herb Adams.
Last edited by gregsdart; 06/15/14 10:35 AM.
8..603 156 mph best, 2905 lbs 549, indy 572-13, alky
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Re: Math majors, trig question!
[Re: gregsdart]
#1633256
06/15/14 11:02 AM
06/15/14 11:02 AM
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Bridgewater, NJ
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Quote:
Quote:
You guys don't have it quite right. .
I am familiar with the rest of the math. All I want is the value that comes from the raised Cg. Because it is variable for each added degree of height, it can't be figured with simple math.
It is the same equation to the speed differential of a U joint. As the angle gets higher, the speed of the driven side of the joint speeds up and slows down in relation to the drive side. That is why simple U joints are always at 90 degrees to each other. The drive shaft itself speeds up and slows down at high angles, but the joints being 90 degrees apart cancel each other, so the output joint ends up as a perfect steady speed match to the input end.
Yeah you can do it either way. I broke the trig triangle into its horizontal and vertical components in that equation above because I figured it would be easier to measure the values that way and I wasn't sure you we're comfortable with sin/cos/tan. (Not being condescending , it's just rare that someone is)
So you have this :
(Engine torque)(tan a) = (weight of car)(tire radius)/(rear gear)(1st gear)(torque converter multiplier)
Where a is not really the angle of wheelie, it's the angle of the the angle of lift of the cg above HORIZONTAL, understanding that even with no wheelie, that angle is not zero.
Also note that in my prior equation, I broke out the car's front end weight and wheel base and this time it is the cars total weight acting at its CG.
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Re: Math majors, trig question!
[Re: Duner]
#1633258
06/15/14 12:24 PM
06/15/14 12:24 PM
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Bridgewater, NJ
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Quote:
How would a person apply this information? and how does knowing the number help you change it?
Honestly, because this is a static analysis and doesn't account for suspension geometry, springs, weight shift, it's of limited value. I guess if you really wanted to see what it would take to get your wheels off the ground in your car, this would get you close, but with proper suspension geometry and weigh transfer, it will take a significantly less torque than what this calculation would suggest.
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Re: Math majors, trig question!
[Re: HT413]
#1633259
06/16/14 01:15 AM
06/16/14 01:15 AM
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gregsdart
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The great thing about math and values is after a guy looks at them for a while and gains some real world on track experience, they begin to make sense in most cases. That is what it is all about to me anyway. I would think a chassis builder that liked math would want to learn things like this to help in deciding how to get close on Cg that works best, based on expected ET and driveline variables. In my particular case, as I mentioned earlier, I wouldn't have set my motor back so far in the chassis I built. It worked great at 9.60s, but became too wheelie prone when more power was added. HT413 thanks for posting the formula  One question; how do you get the value of tangent?
Last edited by gregsdart; 06/16/14 01:24 AM.
8..603 156 mph best, 2905 lbs 549, indy 572-13, alky
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Re: Math majors, trig question!
[Re: gregsdart]
#1633260
06/16/14 03:11 PM
06/16/14 03:11 PM
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HT413
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Sorry, for the long winded response, but I just want to be as clear as I can be...
Finding the tangent of the angle, you just enter the angle value into a calculator and hit the 'tan' button. What tangent means is this... it's just a ratio of the height of the cg divided by the distance from cg to rear contact patch.
So in the attached diagram, say your cg is 18" off the ground and 90" in front of the rear contact patch. Tangent of an angle = opposite side / adjacent side
So tan a = 18/90 = 0.2 Now hit "tan-1" button on a calculator and this gives you an angle of 11.3*. This is without a wheelie, this is just the natural angle of that chassis at rest.
Now, as your front end rises in a wheelie, the horizontal distance from contact patch to cg will get smaller and smaller as it comes closer to passing over the rear wheel. Obviously, the height of use cg will increase as well as it rises.
So, here's the answer to your question (finally!). say you measured an angle of 30* during a wheelie. You would enter that into a calculator and hit "tan". This would spit out a number (0.557). This is the new ratio of height of cg to horizontal distance between cg and rear contact patch. Then enter that into the equation and you're done.
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Re: Math majors, trig question!
[Re: HT413]
#1633261
06/16/14 11:27 PM
06/16/14 11:27 PM
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Joined: May 2008
Posts: 1,131
Thigh-Gap Junction
@#$%&*!
New user name, Same old jerk!
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New user name, Same old jerk!
Joined: May 2008
Posts: 1,131
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Quote:
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Now, as your front end rises in a wheelie, the horizontal distance from contact patch to cg will get smaller and smaller as it comes closer to passing over the rear wheel. Obviously, the height of use cg will increase as well as it rises.
So, here's the answer to your question (finally!). say you measured an angle of 30* during a wheelie. You would enter that into a calculator and hit "tan". This would spit out a number (0.557). This is the new ratio of height of cg to horizontal distance between cg and rear contact patch. Then enter that into the equation and you're done.
I don't see how this relates to the original question:
"If you know the height and position of the CG, it should be possible to calculate just how much force it will take to overcome front end weight without the aid of springs? By that I mean a solid front could be lifted by XX percentage of the actual weight on the front, based on the math."
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Re: Math majors, trig question!
[Re: @#$%&*!]
#1633262
06/17/14 12:35 AM
06/17/14 12:35 AM
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gregsdart
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But if we revert to measuring the horizontal dashed line in the drawing (mine), and divide that by the horizontal distance, it would show a value smaller than the horizontal bottom line.
The Cg will move in an arc rearward, which then puts the tires closer to the vertical position of the Cg. I have to go hunt up my good calculator and try the tangent , then measure for the second method and compare.
I played with the calculator and COS gives the same value as my measurements, or within 1 percent.
Take the number of degrees times COS;
15 degrees x cos = .966
30 degrees x cos = .866
45 degrees x COS = .707
Note that the first 15 degrees the Cg is above horizontal the power to lift the Cg only drops 3.4 percent.
At 30 degrees, it is now down by 13.4 percent
At 45 dgrees, it is down by 29.3 percent.
8..603 156 mph best, 2905 lbs 549, indy 572-13, alky
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Re: Math majors, trig question!
[Re: go green]
#1633266
06/18/14 12:42 PM
06/18/14 12:42 PM
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gregsdart
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Quote:
Are you trying to do a wheelie ? Not really getting where you are going with this .
Actually it is about how to manage a wheelie, or better yet get a car to leave hard without one. I built my own chassis without some of this knowledge, and feel it has some value to those who want to build their own.
For guys determined to not run wheelie bars, it would give them an idea of how much change may be needed to achieve their goals.
Running as fast as you do, your car must be well balanced between torque to the tires, Cg, and front shocks and springs. Looking at the stats for your car (SLR, engine torque, weight distribution, spring and shock rates, and finally, the angle of tire to Cg) would show how close or far away from the ideal other cars with the same ET goals are.
The percentage of change is not great for the first ten to fifteen degrees (less than 4 percent) so this info derived from the Cg to tires angle is mainly aimed at cars in the slower classes.
8..603 156 mph best, 2905 lbs 549, indy 572-13, alky
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