Idea how to carry 2 wheels on the nose
02-11-2005, 05:54 AM
#21
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Originally posted by Da Gimp
Why would you even need to do this? I can't find the thread, but its possible to carry 4 wheels/tires inside the car. (2 in passenger seat, 1 in the far back, and 1 on top of the thing behind the seats)
Why would you even need to do this? I can't find the thread, but its possible to carry 4 wheels/tires inside the car. (2 in passenger seat, 1 in the far back, and 1 on top of the thing behind the seats)
02-11-2005, 06:17 AM
#22
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Here some pictures I did yesterday... In addition to *engeneer's* problem, it's too low clearance to be safe... got to think of some other way...
(don't comment on the car condition
)
(don't comment on the car condition
)
02-11-2005, 08:35 AM
#24
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personally, I would go this route
http://www.harborfreight.com/cpi/cta...emnumber=42708
its small enough to stand up in the garage out of the way when not in use.
http://www.harborfreight.com/cpi/cta...emnumber=42708
its small enough to stand up in the garage out of the way when not in use.
02-11-2005, 09:15 AM
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<- Engineer 3: Question for Engineers 1 and 2:
Explain your idea that there won't be 3000 lb. of tension at the hitch. It seems to me that regardless of where the (backwards) friction forces are in the car, the only forward force comes from the tow/chain at the hitch. Thus, this force has to overcome the sum of these friction forces (which basically are the forces of the distributed weight of the car times the friction coefficient of the road). It doesn't make sense to me that the tow would be towing less...there's no other force to help overcome these "friction forces"
Also, after the initial 3000 lb load is applied, the force required decreases as the car picks up it's own momentum.
I might be wrong, but if I am I'd like to know....good discussion though.
All in all though, this idea makes the car ugly and introduces a helluva lot of drag to the front of the car.
Seriously, why don't you just put them in your passenger seat?
Explain your idea that there won't be 3000 lb. of tension at the hitch. It seems to me that regardless of where the (backwards) friction forces are in the car, the only forward force comes from the tow/chain at the hitch. Thus, this force has to overcome the sum of these friction forces (which basically are the forces of the distributed weight of the car times the friction coefficient of the road). It doesn't make sense to me that the tow would be towing less...there's no other force to help overcome these "friction forces"
Also, after the initial 3000 lb load is applied, the force required decreases as the car picks up it's own momentum.
I might be wrong, but if I am I'd like to know....good discussion though.
All in all though, this idea makes the car ugly and introduces a helluva lot of drag to the front of the car.
Seriously, why don't you just put them in your passenger seat?
02-11-2005, 09:29 AM
#26
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I often have a passenger! And she doesn't feel comfortable with 2 tires on a lap 
And trailer = money + weight + space in the garage... I was trying to come up with some simple solution.
Now, engeneers #1 and #2. What would be the reasonable way to install roof rack on our Z?
Porsches have integrated roof rack installation points... Wouldn't it be great to have them on a Z too...
And trailer = money + weight + space in the garage... I was trying to come up with some simple solution.
Now, engeneers #1 and #2. What would be the reasonable way to install roof rack on our Z?
Porsches have integrated roof rack installation points... Wouldn't it be great to have them on a Z too...
Last edited by Vlad; 02-11-2005 at 09:32 AM.
02-11-2005, 09:36 AM
#27
Originally posted by JThrilla
<- Engineer 3: Question for Engineers 1 and 2:
Explain your idea that there won't be 3000 lb. of tension at the hitch. It seems to me that regardless of where the (backwards) friction forces are in the car, the only forward force comes from the tow/chain at the hitch. Thus, this force has to overcome the sum of these friction forces (which basically are the forces of the distributed weight of the car times the friction coefficient of the road). It doesn't make sense to me that the tow would be towing less...there's no other force to help overcome these "friction forces"
Also, after the initial 3000 lb load is applied, the force required decreases as the car picks up it's own momentum.
I might be wrong, but if I am I'd like to know....good discussion though.
All in all though, this idea makes the car ugly and introduces a helluva lot of drag to the front of the car.
Seriously, why don't you just put them in your passenger seat?
<- Engineer 3: Question for Engineers 1 and 2:
Explain your idea that there won't be 3000 lb. of tension at the hitch. It seems to me that regardless of where the (backwards) friction forces are in the car, the only forward force comes from the tow/chain at the hitch. Thus, this force has to overcome the sum of these friction forces (which basically are the forces of the distributed weight of the car times the friction coefficient of the road). It doesn't make sense to me that the tow would be towing less...there's no other force to help overcome these "friction forces"
Also, after the initial 3000 lb load is applied, the force required decreases as the car picks up it's own momentum.
I might be wrong, but if I am I'd like to know....good discussion though.
All in all though, this idea makes the car ugly and introduces a helluva lot of drag to the front of the car.
Seriously, why don't you just put them in your passenger seat?
Just as you mentioned above "Thus, this force has to overcome the sum of these friction forces (which basically are the forces of the distributed weight of the car times the friction coefficient of the road)." is absolutely correct.
Let's assume:
- Car = 3000lbs
- Static friction force between ground and tire is 1 (meaning that tires will roll and not slide on surface)
- static friction force of axle, gears, etc. is 0.4
Then 3000lbs x 0.4 = 1200lb-f of tension.
Therefore, it is only a fraction of the car's weight. Whatever the exact coefficient of friction is, I don't know, but it's less than 1. To say that there is 3000lbs of tension is saying that all friction forces = 1. i.e. 3000lbs x 1 = 3000lb-f meaning everything is rigid.
02-11-2005, 09:49 AM
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Ahhh, you're right. The more I thought about this after I made that post, I realized that a coefficient of friction (mu, isn't it?) can't be greater than 1.
I see what you're saying...thanks for the clarification. I guess my question now is (I'm learning here)...how can you only take into the consideration of the friction caused by the axle, gears, etc. and not that of the road/tires? Shouldn't there be some "average" coeff. value that incorporates all of these factors? This is interesting...in school we always treat things as perfectly rigid, etc. to simplify the problem.
btw, what kind of engineer are you? Thanks for your help.
I see what you're saying...thanks for the clarification. I guess my question now is (I'm learning here)...how can you only take into the consideration of the friction caused by the axle, gears, etc. and not that of the road/tires? Shouldn't there be some "average" coeff. value that incorporates all of these factors? This is interesting...in school we always treat things as perfectly rigid, etc. to simplify the problem.
btw, what kind of engineer are you? Thanks for your help.
02-11-2005, 10:05 AM
#29
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trailer can be tipped on it's back and stored. Can also carry tools etc. CAn use trailer for other transport needs too. Hitch receive can be made to be removed. (made a hitch for my 1972 Z back in 72!). These small trailers are what the Miata racers use. cb
02-11-2005, 12:08 PM
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that design is no good if you hit a bump in the road or go up a slight incline, the tire will catch on the ground... you are extending your front bumper which makes it more sensitive to slight angles to the roadway...if that makes any sense?
02-11-2005, 05:26 PM
#31
Originally posted by Vlad
After some thinking... The piece of aluminum for a tow hook is welded UNDER the aluminum bumper. When car is towed, forces are trying to shift it from the bumper (A). With this nose tire rack, force will be trying to pull it from the bumper (B)
So with some common sence applied (my kind), connection should be stronger in B direction than in A. Think of two strong magnets, and how to separte them.
Now, the question is, how bumper is attached to the frame, and how those connections will hold?
After some thinking... The piece of aluminum for a tow hook is welded UNDER the aluminum bumper. When car is towed, forces are trying to shift it from the bumper (A). With this nose tire rack, force will be trying to pull it from the bumper (B)
So with some common sence applied (my kind), connection should be stronger in B direction than in A. Think of two strong magnets, and how to separte them.
Now, the question is, how bumper is attached to the frame, and how those connections will hold?
02-12-2005, 10:14 PM
#32
Originally posted by JThrilla
Ahhh, you're right. The more I thought about this after I made that post, I realized that a coefficient of friction (mu, isn't it?) can't be greater than 1.
I see what you're saying...thanks for the clarification. I guess my question now is (I'm learning here)...how can you only take into the consideration of the friction caused by the axle, gears, etc. and not that of the road/tires? Shouldn't there be some "average" coeff. value that incorporates all of these factors? This is interesting...in school we always treat things as perfectly rigid, etc. to simplify the problem.
btw, what kind of engineer are you? Thanks for your help.
Ahhh, you're right. The more I thought about this after I made that post, I realized that a coefficient of friction (mu, isn't it?) can't be greater than 1.
I see what you're saying...thanks for the clarification. I guess my question now is (I'm learning here)...how can you only take into the consideration of the friction caused by the axle, gears, etc. and not that of the road/tires? Shouldn't there be some "average" coeff. value that incorporates all of these factors? This is interesting...in school we always treat things as perfectly rigid, etc. to simplify the problem.
btw, what kind of engineer are you? Thanks for your help.
From the picture, you see a conveyor belt. Same concept, basically, each component will be multiplied by its respective friction force in the system and summed together to obtain the force necessary to two the car. Not really an average, but more a summation of each force in the system that prevents it from moving. Also notice that it equals m*a and not zero, b/c this is not a static problem, its a dynamic one b/c the car moves. Another thing not taken into account is inertia(the force that resists motion) of the components in the car. Anyways, hope that helps.
02-13-2005, 04:48 PM
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Originally posted by engiNERD350Z
<------ Engineer #2's answer.
Just as you mentioned above "Thus, this force has to overcome the sum of these friction forces (which basically are the forces of the distributed weight of the car times the friction coefficient of the road)." is absolutely correct.
Let's assume:
- Car = 3000lbs
- Static friction force between ground and tire is 1 (meaning that tires will roll and not slide on surface)
- static friction force of axle, gears, etc. is 0.4
Then 3000lbs x 0.4 = 1200lb-f of tension.
Therefore, it is only a fraction of the car's weight. Whatever the exact coefficient of friction is, I don't know, but it's less than 1. To say that there is 3000lbs of tension is saying that all friction forces = 1. i.e. 3000lbs x 1 = 3000lb-f meaning everything is rigid.
<------ Engineer #2's answer.
Just as you mentioned above "Thus, this force has to overcome the sum of these friction forces (which basically are the forces of the distributed weight of the car times the friction coefficient of the road)." is absolutely correct.
Let's assume:
- Car = 3000lbs
- Static friction force between ground and tire is 1 (meaning that tires will roll and not slide on surface)
- static friction force of axle, gears, etc. is 0.4
Then 3000lbs x 0.4 = 1200lb-f of tension.
Therefore, it is only a fraction of the car's weight. Whatever the exact coefficient of friction is, I don't know, but it's less than 1. To say that there is 3000lbs of tension is saying that all friction forces = 1. i.e. 3000lbs x 1 = 3000lb-f meaning everything is rigid.
I'm pretty sure you could hang the car from the ceiling by that tow hook (or pretty close). lol
Last edited by Tex Willer; 02-13-2005 at 04:52 PM.
02-15-2005, 05:15 AM
#36
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another thing is...you see how long your front end will be??
just picture theres a driveway you have 2 clear with a f-ed up angle..where you will barely clear with stock front end anyway since the z sits so low...adding inches in length is not an option at all...you wont be clearing anything...i dont know how else 2 describe it...xcuze my poor vocabulary...im german=/...lol
just picture theres a driveway you have 2 clear with a f-ed up angle..where you will barely clear with stock front end anyway since the z sits so low...adding inches in length is not an option at all...you wont be clearing anything...i dont know how else 2 describe it...xcuze my poor vocabulary...im german=/...lol
02-16-2005, 10:48 AM
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yeah..i was thinking just throw the tires on if it was closer..but in that case..guess not...but yeah..having them on the front is not an option...hmm..the inside is like the only option i see...unless you wanna make a custom rack on top or on the rear of your Z...i say take the passenger seat out..since your goin 2 the track...put them there...secure them...and when ur on the track it'll be less weight also since the removed seat...sounds good?