How much does car boot/trunk weight increase petrol costs?

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Discussion Overview

The discussion centers on the impact of carrying extra weight, specifically golf clubs, in a car on fuel costs. Participants explore various factors affecting fuel efficiency, including kinetic energy, rolling resistance, and other influences on gas mileage.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant queries the cost implications of leaving golf clubs in their Hyundai i30, suggesting a desire for calculations based on average car specifications.
  • Another participant estimates a potential fuel saving of about 0.2% in urban driving conditions, noting that this is less than the savings from reducing fuel tank levels.
  • Discussion includes the concept that extra mass requires more kinetic energy during acceleration, which may lead to increased fuel consumption unless the vehicle has regenerative braking.
  • It is mentioned that additional weight increases rolling resistance, although a 10kg increase may not significantly impact fuel efficiency.
  • A participant introduces the idea that driving over a significant elevation, such as a 1000-foot ridge, adds to the energy expenditure due to potential energy changes.
  • Three factors affecting gas mileage are identified: wind resistance, friction from various components, and the energy required to accelerate additional mass, with the latter being emphasized as a significant contributor.

Areas of Agreement / Disagreement

Participants express varying views on the significance of extra weight on fuel costs, with some suggesting minimal impact while others highlight multiple contributing factors. The discussion remains unresolved regarding the exact quantification of these effects.

Contextual Notes

Participants acknowledge that the calculations depend on various assumptions, including vehicle specifications, driving conditions, and the interplay of different factors affecting fuel efficiency.

Who May Find This Useful

Individuals interested in automotive efficiency, physics of motion, and practical implications of vehicle load on fuel consumption may find this discussion relevant.

mtallyn
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I often leave my golf clubs in the car rather than bothering to take them out I just wondered how much extra this is costing me. I have an Hyundai i30 ps109, ~1.2-1.3 tonnes and the clubs are about 10kg but more than happy for average car specs/approximations to be used. Can anyone do the maths? I'd be really curious to know.
 
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You'd be lucky to get about 0.2% saving around town, less than half that on the motorway/autoroute/bahn/strada/pista etc. (www.drivealuminum.org/research-resources/.../2008-Ricardo-Study.pdf ). Note that this is probably less than the weight of a quarter of a tank of fuel, so you would save more by only filling up to half way when refuelling.
 
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There are (at least) two separate parts to this.

First, every time you accelerate, the extra mass acquires kinetic energy (##mv^2/2##), and unless you car has a regenerative braking system, your brakes convert that energy into heat when you slow down.

Second, unless you adjust the tire pressures, extra weight increases the rolling resistance of the tires.

An extra 10kg weight probably won't make much difference, though.
 
Thirdly; If you play golf once per week, but drive to work every day over a 1000 foot ridge. Then the mass must be raised 1000 ft, twice each day. PE = m * g * h
 
There are three factors (maybe more) that affect gas mileage.

The first is wind resistance. This could be the biggest factor but the golf clubs won't affect it.

The second is the friction generated by the engine, air conditioning, bearings etc. The clubs won't appreciably affect this either.

The third is as has been mentioned already, the increased energy expended to accelerate the extra mass and the increased rolling resistance. This factor is probably proportional to the mass of the car, thus the ratio of the mass of the car without the clubs to the mass with the clubs would be the multiplier for this third factor. What is left is to determine what proportion of the total is the third factor.
 

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